a) \(x^2-36=0\)

\(\Leftrightarrow x^2-6^2=0\)

\(\Leftrightarrow\left(x-6\right)\left(x+6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-6=0\\x+6=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-6\end{matrix}\right.\) 

Vậy: ... 

b) \(x^2-10x+25=0\)

\(\Leftrightarrow x^2-2\cdot x\cdot5+5^2=0\)

\(\Leftrightarrow\left(x-5\right)^2=0\)

\(\Leftrightarrow x-5=0\)

\(\Leftrightarrow x=5\)

Vậy: ... 

a) \(x^2-36=0\)

\(\Leftrightarrow x^2=36\)

\(\Leftrightarrow x^2=\left(\pm6\right)^2\)

\(\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-6\end{matrix}\right.\)

Vậy \(x\in\left\{6;-6\right\}\)

b) \(x^2-10x+25=0\)

\(\Leftrightarrow x^2-2.x.5+5^2=0\)

\(\Leftrightarrow\left(x-5\right)^2=0\)

\(\Leftrightarrow x-5=0\)

\(\Leftrightarrow x=5\)

Vậy \(x=5\)

a) \(\left(3x-1\right)\left(x+2\right)-\left(x+2\right)^2\)

\(=\left(3x^2+6x-x-2\right)-\left(x+2\right)^2\)

\(=\left(3x^2+5x-2\right)-\left(x^2+4x+4\right)\)

\(=3x^2+5x-2-x^2-4x-4\)

\(=2x^2+x-6\) 

b) \(\left(x-1\right)\left(x+1\right)-\left(x^2-2x+1\right)\)

\(=\left(x^2-1\right)-\left(x^2-2x+1\right)\) 

\(=x^2-1-x^2+2x-1\)

\(=2x-2\)

c) \(\left(x-4\right)\left(4+x\right)+2x\left(x-3\right)\)

\(=\left(x-4\right)\left(x+4\right)+2x\left(x-3\right)\)

\(=\left(x^2-16\right)+2x^2-6x\)

\(=x^2-16+2x^2-6x\)

\(=3x^2-6x-16\)

d) \(\left(x-1\right)\left(x^2-1\right)+\left(x+2\right)^3\)

\(=\left(x^3-x-x^2+1\right)+\left(x^3+6x^2+12x+8\right)\)

\(=x^3-x-x^2+1+x^3+6x^2+12x+8\)

\(=2x^3+5x^2+11x+9\)

e) \(\left(2x-1\right)^2-\left(2x-5\right)\left(x+5\right)\)

\(=\left(4x^2-4x+1\right)-\left(2x^2+10x-5x-25\right)\)

\(=\left(4x^2-4x+1\right)-\left(2x^2+5x-25\right)\)

\(=4x^2-4x+1-2x^2-5x+25\)

\(=2x^2-9x+26\)

f) \(\left(3x+1\right)^2-\left(x^2-1\right)\left(x^2+2\right)\)

\(=\left(9x^2+6x+1\right)-\left(x^4+2x^2-x^2-2\right)\)

\(=\left(9x^2+6x+1\right)-\left(x^4+x^2-2\right)\)

\(=9x^2+6x+1-x^4-x^2+2\)

\(=-x^4+8x^2+6x+3\) 

g) \(\left(x^2+1\right)^2-\left(x^2-1\right)\left(x^2+2\right)\)

\(=\left(x^4+2x^2+1\right)-\left(x^4+2x^2-x^2-2\right)\)

\(=\left(x^4+2x^2+1\right)-\left(x^4+x^2-2\right)\)

\(=x^4+2x^2+1-x^4-x^2+2\)

\(=x^2+3\)

h) \(\left(2x^2-4\right)^2-\left(2x^2+4\right)^2\)

\(=\left(4x^4-16x^2+16\right)-\left(4x^4+16x^2+16\right)\)

\(=4x^4-16x^2+16-4x^4-16x^2-16\)

\(=-32x^2\)

   ∆ABC có BE là đường phân giác (gt)

  ∆ABC vuông tại A (gt)

⇒ BC² = AB² + AC² (Pythagore)

⇒ BC² - AB² = AC²

= (3 + 5)²

= 64

Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:

  Ta có:

BC² = AB² + AC² (Pythagore)

= 6² + 64

= 100

⇒ BC = 10

vì BE là đường phân giác của tam giác ABC nên ta có:

\(\dfrac{AE}{EC}=\dfrac{AB}{BC}=\dfrac{3}{5}\)

\(BC=\dfrac{5}{3}AB\)

áp dụng định lý pythagore vào tam giác ABC ta được:

\(AC^2=AB^2+BC^2\)

tổng độ dài đoạn AC là: 3 + 5 = 8

\(AB^2+BC^2=8^2\\ AB^2+\left(\dfrac{5}{3}AB\right)^2=64\\ AB^2+\dfrac{25}{9}AB^2=64\\ AB^2\cdot\left(1+\dfrac{25}{9}\right)=64\\ AB^2\cdot\dfrac{34}{9}=64\\ AB^2=64:\dfrac{34}{9}=64\cdot\dfrac{9}{34}\\ AB^2=\dfrac{576}{34}\\ AB=\sqrt{\dfrac{576}{34}}\text{≈}4,11\)

độ dài đoạn BC là:

BC² = AC² - AB²

BC² = 64 - 16,8921

BC² = 47,1079

BC = \(\sqrt{47,1079}\) ≈ 6,86

VẬY AB = 4,11; BC  =6,86

Đa thức $2x^4-21x^2+1$ không phân tích thành nhân tử bạn nhé.

Xét ΔABC vuông tại A có \(sinACB=\dfrac{AB}{BC}\)

=>\(\dfrac{5}{BC}=sin30=\dfrac{1}{2}\)

=>BC=10(cm)

ΔABC vuông tại A

=>\(AB^2+AC^2=BC^2\)

=>\(AC=\sqrt{10^2-5^2}=5\sqrt{3}\left(cm\right)\)

Lời giải:

$A=(4a^2+4b^2+c^2+8ab-4ac-4bc)+(4b^2+4c^2+a^2+8bc-4ab-4ac)+(4c^2+4a^2+b^2+8ac-4bc-4ab)$

$=6(a^2+b^2+c^2)=6m$

\(\dfrac{7^{49}-7^{48}}{7^{48}}\)

\(=\dfrac{7^{49}}{7^{48}}-\dfrac{7^{48}}{7^{48}}\)

\(=7-1\)

\(=6\)

\(\dfrac{7^{49}-7^{48}}{7^{48}}\)

\(=\dfrac{7^{48}\cdot7-7^{48}}{7^{48}}\)

\(=\dfrac{7^{48}\left(7-1\right)}{7^{48}}\)

\(=7-1\)

\(=6\)

13) 

a) \(\left\{{}\begin{matrix}7x+4y=2\\5x-2y=16\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}7x+4y=2\\10x-4y=32\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}7x+4y=2\\17x=34\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}7\cdot2+4y=2\\x=2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}4y=2-14\\x=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4y=-12\\x=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-3\\x=2\end{matrix}\right.\)

Vậy: .... 

b) \(\left\{{}\begin{matrix}2x+3y=19\\3x+4y=-14\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}6x+9y=57\\6x+8y=-28\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x+3y=19\\y=85\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x+3\cdot85=19\\y=85\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x=19-255\\y=85\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=-236\\y=85\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-118\\y=85\end{matrix}\right.\)

Vậy: .... 

c) \(\left\{{}\begin{matrix}2x+2y=3\\3x-2y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5x=5\\3x-2y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\3\cdot1-2y=2\end{matrix}\right.\)

 \(\Leftrightarrow\left\{{}\begin{matrix}x=1\\-2y=2-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\-2y=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=\dfrac{1}{2}\end{matrix}\right.\)

Vậy: ....

15) 

a) \(\left\{{}\begin{matrix}5\left(x+2\right)=2\left(y+7\right)\\3\left(x+y\right)=17-x\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}5x+10=2y+14\\3x+3y=17-x\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}5x-2y=14-10\\3x+3y+x=17\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}5x-2y=4\\4x+3y=17\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}15x-6y=12\\8x+6y=34\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}5x-2y=4\\23x=46\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}5\cdot2-2y=4\\x=2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2y=6\\x=2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y=3\\x=2\end{matrix}\right.\)

vậy: ... 

Do đồ thị của hàm số cắt trục hoành tại điểm có hoành độ bằng -2,5 nên đi qua điểm (-2,5; 0)

Thay tọa độ điểm (-2,5; 0) vào hàm số, ta có:

2.(-2,5) + b = 0

-5 + b = 0

b = 0 + 5

b = 5

Vậy hàm số cần xác định là: y = 2x + 5