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4x(x2-2x+3)-3x(x+1)(x-2)+(2x-5)(x-7)-4x(x2-x+1)
=4x3-8x2+12x-3x(x2-2x+x-2)+2x2-14x-5x+35-4x3+4x2-4x
=4x3-8x2+12x-3x3+6x2-3x2+6x+2x2-14x-5x+35-4x3+4x2-4x
=-3x3+x2-5x+35
#H
a,
Ta có: \(a\left(b+1\right)b\left(a+1\right)=\left(a+1\right)\left(b+1\right)\)
\(\Rightarrow ab=\left(a+1\right)\left(b+1\right):\left(a+1\right)\left(b+1\right)=1\)
=>đpcm
b,
Ta có: \(2\left(a+1\right)\left(a+b\right)=\left(a+b\right)\left(a+b+2\right)\)
\(\Rightarrow2a+2=a+b+2\)
\(\Rightarrow a-b=0\)
\(\Rightarrow a^2+b^2=2ab\)
\(\Rightarrow a^2+b^2=2\) (đpcm)
1.
a. x3 - 4x2 - xy2 + 4x
= x ( x2 - 4x + 4 - y2 )
= x [ ( x - 2 )2 - y2 ]
= x ( x - y - 2 ) ( x + y - 2 )
b. x2 - x - 2 = x2 + x - 2x - 2 = x ( x + 1 ) - 2 ( x + 1 ) = ( x - 2 ) ( x + 1 )
c. x4 + 4
= ( x4 + 2x3 + 2x2 ) - ( 2x3 + 4x2 + 4x ) + ( 2x2 + 4x + 4 )
= x2 ( x2 + 2x + 2 ) - 2x ( x2 + 2x + 2 ) + 2 ( x2 + 2x + 2 )
= ( x2 + 2x + 2 ) ( x2 - 2x + 2 )
(x2 - 3x)(-5x2 + x) - x2(x - 7) + 5x2(x - 2)(x - 4) - (x + 1)(x + 3)
= -5x4 + 16x3 - 3x2 - x3 + 7x2 + 5x2(x2 - 6x + 8) - x2 - 4x - 3
= -5x4 + 15x3 + 3x2 - 4x - 3 + 5x4 - 30x3 + 40x2
= -15x3 + 43x2 - 4x - 3
(5x - 2)(x + 1) - 3x(x2 - x - 3) - 2x(x - 5)(x + 4) + 5x(x2 - 7)
= 5x2 + 3x - 2 - 3x3 + 3x2 + 9x - 2x(x2 - x - 20) + 5x3 - 35x
= 2x3 + 8x2 - 23x - 2 - 2x3 + 2x2 + 40x = 10x2 + 17x - 2
a) \(\left(\right. x^{2} + 3 x + 2 \left.\right) \left(\right. x^{2} + 3 x + 3 \left.\right) - 2 = 0\)
Đặt
\(t = x^{2} + 3 x + 2.\)Khi đó:
\(x^{2} + 3 x + 3 = t + 1.\)Phương trình trở thành:
\(t \left(\right. t + 1 \left.\right) - 2 = 0\) \(t^{2} + t - 2 = 0\) \(\left(\right. t + 2 \left.\right) \left(\right. t - 1 \left.\right) = 0.\)TH1: \(t = 1\)
\(x^{2} + 3 x + 2 = 1\) \(x^{2} + 3 x + 1 = 0\) \(x = \frac{- 3 \pm \sqrt{5}}{2} .\)TH2: \(t = - 2\)
\(x^{2} + 3 x + 2 = - 2\) \(x^{2} + 3 x + 4 = 0\)\(\Delta < 0\), vô nghiệm.
Đáp số:
\(\boxed{x = \frac{- 3 \pm \sqrt{5}}{2}} .\)b) \(\left(\right. x + 1 \left.\right) \left(\right. x + 2 \left.\right) \left(\right. x + 3 \left.\right) \left(\right. x + 4 \left.\right) - 24 = 0\)
Ta có:
\(\left(\right. x + 1 \left.\right) \left(\right. x + 4 \left.\right) = x^{2} + 5 x + 4 ,\) \(\left(\right. x + 2 \left.\right) \left(\right. x + 3 \left.\right) = x^{2} + 5 x + 6.\)Đặt
\(t = x^{2} + 5 x + 5.\)Khi đó:
\(x^{2} + 5 x + 4 = t - 1 , x^{2} + 5 x + 6 = t + 1.\)Phương trình:
\(\left(\right. t - 1 \left.\right) \left(\right. t + 1 \left.\right) - 24 = 0\) \(t^{2} - 25 = 0\) \(t = \pm 5.\)TH1: \(t = 5\)
\(x^{2} + 5 x = 0\) \(x \left(\right. x + 5 \left.\right) = 0\) \(x = 0 , \&\text{nbsp}; - 5.\)TH2: \(t = - 5\)
\(x^{2} + 5 x + 10 = 0\)\(\Delta < 0\), vô nghiệm.
Đáp số:
\(\boxed{x = 0 ; \textrm{ } - 5.}\)c) \(\left(\right. x^{2} + 5 x \left.\right)^{2} - 2 \left(\right. x^{2} + 5 x \left.\right) = 24\)
Đặt
\(t = x^{2} + 5 x .\)Ta có:
\(t^{2} - 2 t - 24 = 0\) \(\left(\right. t - 6 \left.\right) \left(\right. t + 4 \left.\right) = 0.\)TH1: \(t = 6\)
\(x^{2} + 5 x - 6 = 0\) \(\left(\right. x + 6 \left.\right) \left(\right. x - 1 \left.\right) = 0\) \(x = - 6 , \&\text{nbsp}; 1.\)TH2: \(t = - 4\)
\(x^{2} + 5 x + 4 = 0\) \(\left(\right. x + 1 \left.\right) \left(\right. x + 4 \left.\right) = 0\) \(x = - 1 , \&\text{nbsp}; - 4.\)Đáp số:
\(\boxed{x=-6,;-4,;-1;1.}\)d) \(x \left(\right. x + 1 \left.\right) \left(\right. x - 1 \left.\right) \left(\right. x + 2 \left.\right) = 24\)
Ta có:
\(x \left(\right. x + 1 \left.\right) = x^{2} + x ,\) \(\left(\right. x - 1 \left.\right) \left(\right. x + 2 \left.\right) = x^{2} + x - 2.\)Đặt
\(t = x^{2} + x .\)Phương trình:
\(t \left(\right. t - 2 \left.\right) = 24\) \(t^{2} - 2 t - 24 = 0\) \(\left(\right. t - 6 \left.\right) \left(\right. t + 4 \left.\right) = 0.\)TH1: \(t = 6\)
\(x^{2} + x - 6 = 0\) \(\left(\right. x + 3 \left.\right) \left(\right. x - 2 \left.\right) = 0\) \(x = - 3 , \&\text{nbsp}; 2.\)TH2: \(t = - 4\)
\(x^{2} + x + 4 = 0\)\(\Delta < 0\), vô nghiệm.
Đáp số:
\(\boxed{x=-3;2.}\)e) \(\left(\right. x^{2} - x \left.\right)^{2} - 2 = x^{2} - x\)
Đặt
\(t = x^{2} - x .\)Ta được:
\(t^{2} - t - 2 = 0\) \(\left(\right. t - 2 \left.\right) \left(\right. t + 1 \left.\right) = 0.\)TH1: \(t = 2\)
\(x^{2} - x - 2 = 0\) \(\left(\right. x - 2 \left.\right) \left(\right. x + 1 \left.\right) = 0\) \(x = 2 , \&\text{nbsp}; - 1.\)TH2: \(t = - 1\)
\(x^{2} - x + 1 = 0\)\(\Delta < 0\), vô nghiệm.
Đáp số:
\(\boxed{x=-1;2.}\)a: \(\left(x^2+3x+2\right)\left(x^2+3x+3\right)-2=0\)
=>\(\left(x^2+3x\right)^2+5\left(x^2+3x\right)+6-2=0\)
=>\(\left(x^2+3x\right)^2+5\left(x^2+3x\right)+4=0\)
=>\(\left(x^2+3x+4\right)\left(x^2+3x+1\right)=0\)
mà \(x^2+3x+4=\left(x+\frac32\right)^2+\frac74>0\forall x\)
nên \(x^2+3x+1=0\)
=>\(x^2+3x+\frac94=\frac54\)
=>\(\left(x+\frac32\right)^2=\frac54\)
=>\(\left[\begin{array}{l}x+\frac32=\frac{\sqrt5}{2}\\ x+\frac32=-\frac{\sqrt5}{2}\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac{\sqrt5-3}{2}\\ x=\frac{-\sqrt5-3}{2}\end{array}\right.\)
b: \(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-24=0\)
=>\(\left(x^2+5x+4\right)\left(x^2+5x+6\right)-24=0\)
=>\(\left(x^2+5x+5\right)^2-1-24=0\)
=>\(\left(x^2+5x+5\right)^2-25=0\)
=>\(\left(x^2+5x+5-5\right)\left(x^2+5x+5+5\right)=0\)
=>\(\left(x^2+5x\right)\left(x^2+5x+10\right)=0\)
mà \(x^2+5x+10=\left(x+\frac52\right)^2+\frac{15}{4}>0\forall x\)
nên \(x^2+5x=0\)
=>x(x+5)=0
=>x=0 hoặc x=-5
c: \(\left(x^2+5x\right)^2-2\left(x^2+5x\right)=24\)
=>\(\left(x^2+5x\right)^2-2\left(x^2+5x\right)-24=0\)
=>\(\left(x^2+5x-6\right)\left(x^2+5x+4\right)=0\)
=>(x+6)(x-1)(x+1)(x+4)=0
=>x∈{-6;1;-1;-4}
d: x(x+1)(x-1)(x+2)=24
=>\(\left(x^2+x\right)\left(x^2+x-2\right)=24\)
=>\(\left(x^2+x\right)^2-2\left(x^2+x\right)-24=0\)
=>\(\left(x^2+x-6\right)\left(x^2+x+4\right)=0\)
mà \(x^2+x+4=\left(x+\frac12\right)^2+\frac{15}{4}>0\forall x\)
nên \(x^2+x-6=0\)
=>(x+3)(x-2)=0
=>\(\left[\begin{array}{l}x+3=0\\ x-2=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=-3\\ x=2\end{array}\right.\)
e: \(\left(x_{}^2-x\right)^2-2=x^2-x\)
=>\(\left(x^2-x\right)^2-\left(x^2-x\right)-2=0\)
=>\(\left(x^2-x-2\right)\left(x^2-x+1\right)=0\)
mà \(x^2-x+1=\left(x-\frac12\right)^2+\frac34\ge\frac34>0\forall x\)
nên \(x^2-x-2=0\)
=>(x-2)(x+1)=0
=>x∈{2;-1}
dài thế