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a) \(\left(3x-5\right)\left(2x+3\right)-\left(2x-3\right)\left(3x+7\right)-2x\left(x-4\right)\)
\(=\left(6x^2-x-15\right)-\left(6x^2+5x-21\right)-\left(2x^2-8x\right)\)
\(=6x^2-x-15-6x^2-5x+21-2x^2+8x\)
\(=-2x^2+2x+6\)
\(=-2\left(x^2-x-3\right)\)
b) \(\left(x^2+2\right)^2-\left(x+2\right)\left(x-2\right)\left(x^2+4\right)\)
\(=\left(x^2+2\right)^2-\left(x^2-4\right)\left(x^2+4\right)\)
\(=\left(x^2+2\right)^2-\left(x^4-16\right)\)
\(=\left(x^4+4x^2+4\right)-\left(x^4-16\right)\)
\(=x^4+4x^2+4-x^4+16\)
\(=4x^2+20\)
\(=4\left(x^2+5\right)\)
c) \(\left(2x-y\right)^2-2\left(x+3y\right)^2-\left(1+3x\right)\left(3x-1\right)\)
\(=\left(4x^2-4xy+y^2\right)-2\left(x^2+6xy+9y^2\right)-\left(9x^2-1\right)\)
\(=4x^2-4xy+y^2-2x^2-16xy-18y^2-9x^2+1\)
\(=-7x^2-20xy-17y^2+1\)
d) \(\left(x^2-1\right)^3-\left(x^4+x^2+1\right)\left(x^2-1\right)\)
\(=\left(x^6-3x^4+3x^2-1\right)-\left(x^6-1\right)\)
\(=x^6-3x^4+3x^2-1-x^6+1\)
\(=-3x^4+3x^2\)
\(=-3x^2\left(x^2-1\right)\)
\(=-3x^2\left(x-1\right)\left(x+1\right)\)
e) \(\left(2x-1\right)^2-2\left(4x^2-1\right)+\left(2x+1\right)^2\)
\(=\left(2x-1\right)^2-2\left(2x-1\right)\left(2x+1\right)+\left(2x+1\right)^2\)
\(=\left[\left(2x-1\right)-\left(2x+1\right)\right]^2\)
\(=\left(2x-1-2x-1\right)^2\)
\(=\left(-2\right)^2=4\)
g) \(\left(x-y+z\right)^2+\left(y-z\right)^2-2\left(x-y+z\right)\left(z-y\right)\)
\(=\left(x-y+z\right)^2+2\left(x-y+z\right)\left(y-z\right)+\left(y-z\right)^2\)
\(=\left(x-y+z+y+z\right)^2\)
\(=\left(x+2z\right)^2\)
h) \(\left(2x+3\right)^2+\left(2x+5\right)^2-\left(4x+6\right)\left(2x+5\right)\)
\(=\left(2x+3\right)^2-2\left(2x+3\right)\left(2x+5\right)+\left(2x+5\right)^2\)
\(=\left[\left(2x+3\right)-\left(2x+5\right)\right]^2\)
\(=\left(2x+3-2x-5\right)^2\)
\(=\left(-2\right)^2=4\)
i) \(5x^2-\dfrac{10x^3+15x^2-5x}{-5x}-3\left(x+1\right)\)
\(=5x^2-\dfrac{-5x\left(-2x^2-3x+1\right)}{-5x}-3\left(x+1\right)\)
\(=5x^2-\left(-2x^2-3x+1\right)-3\left(x+1\right)\)
\(=5x^2+2x^2+3x-1-3x-3\)
\(=7x^2-4\)
Bạn làm bài kiểm tra hả sao nhiều bài tek. Mk làm mất khá nhiều tg luôn đó 
Có một số câu thì mình không làm được. Mong bạn thông cảm!!!
Bài 1:
a. A = x^2 - 5x - 1
\(=x^2-5x+\frac{25}{4}-\frac{29}{4}\)
\(=x^2-5x+\left(\frac{5}{2}\right)^2-\frac{29}{4}\)
\(=\left(x-\frac{5}{2}\right)^2-\frac{29}{4}\ge0-\frac{29}{4}=-\frac{29}{4}\)
Dấu = khi x=5/2
Vậy MinC=-29/4 khi x=5/2
2. Tìm x:
a. ( 2x - 3 )^2 - ( 4x + 1 )( 4x - 1 ) = ( 2x - 1 ).( 3 - 7x )
=>4x2-12x+9+1-16x2=-14x2+13x-3
=>-12x2-12x+10=-14x2+13x-3
=>2x2-25x+13=0
\(\Rightarrow2\left(x-\frac{25}{4}\right)^2-\frac{521}{8}=0\)
\(\Rightarrow\left(x-\frac{25}{4}\right)^2=\frac{521}{16}\)
\(\Rightarrow x-\frac{25}{4}=\pm\sqrt{\frac{521}{16}}\)
\(\Rightarrow x=\frac{25}{4}\pm\frac{\sqrt{521}}{4}\)
c. 4.( x - 3 ) - ( x + 2 ) = 0
=>4x-12-x-2=0
=>3x-14=0
=>3x=14
=>x=14/3
a) 3( x - y ) - 5x( y - x )
= 3( x - y ) - 5x[ -( x - y ) ]
= 3( x - y ) + 5x( x - y )
= ( 3 + 5x )( x - y )
b) x3 + 2x2y + xy2 - 9x
= x( x2 + 2xy + y2 - 9 )
= x[ ( x + y )2 - 32 ]
= x( x + y - 3 )( x + y + 3 )
c) 14x2y - 21xy2 + 28x2y2
= 7xy( 2x - 3y + 4xy )
Bài giải
\(a,\text{ }3\left(x-y\right)-5x\left(y-x\right)\)
\(=3\left(x-y\right)+5x\left(x-y\right)\)
\(=\left(x-y\right)\left(3+5x\right)\)
\(b,\text{ }x^3+2x^2y+xy^2-9x\)
\(=x\left(x^2+2xy+y^2-9\right)\)
\(=x\left[\left(x+y\right)^2-3^2\right]\)
\(=x\left(x+y+3\right)\left(x+y-3\right)\)
\(c,\text{ }14x^2y-21xy^2+28x^2y\)
\(=7xy\left(2x-3y+4x\right)\)
\(=7xy\left(6x-3y\right)\)
a)
\(25x^2-9(x+y)^2=(5x)^2-(3x+3y)^2\)
\(=(5x-3x-3y)(5x+3x+3y)=(2x-3y)(8x+3y)\)
b)
\(x^2-x-2=x^2+x-2x-2=x(x+1)-2(x+1)=(x-2)(x+1)\)
c)
\(3x^2-11x+6=3x^2-9x-2x+6\)
\(=3x(x-3)-2(x-3)=(x-3)(3x-2)\)
d)
\(x^2+5x+8\): biểu thức không phân tích được thành nhân tử
e)
\(x^2+8x+7=x^2+x+7x+7\)
\(=x(x+1)+7(x+1)=(x+1)(x+7)\)
g)
\(x^2-6x-16=x^2-6x+9-25\)
\(=(x-3)^2-5^2=(x-3-5)(x-2+5)=(x-8)(x+2)\)
h)
\(4x^2-8x+3=4(x^2-2x+1)-1\)
\(=4(x-1)^2-1=(2x-2)^2-1^2=(2x-2-1)(2x-2+1)\)
\(=(2x-3)(2x-1)\)
i)
\(3x^2-11x+6=3x^2-9x-2x+6\)
\(=3x(x-3)-2(x-3)=(3x-2)(x-3)\)
a) Ta có: \(\frac{\left(2x+1\right)^2}{5}-\frac{\left(x-1\right)^2}{3}=\frac{7x^2-14x-5}{15}\)
\(\Leftrightarrow\frac{\left(2x+1\right)^2\cdot3}{15}-\frac{5\left(x-1\right)^2}{15}-\frac{7x^2-14x-5}{15}=0\)
\(\Leftrightarrow3\left(4x^2+4x+1\right)-5\left(x^2-2x+1\right)-7x^2+14x+5=0\)
\(\Leftrightarrow12x^2+12x+3-5x^2+10x-5-7x^2+14x+5=0\)
\(\Leftrightarrow36x+3=0\)
\(\Leftrightarrow36x=-3\)
\(\Leftrightarrow x=\frac{-3}{36}\)
Vậy: \(x=\frac{-3}{36}\)
b) Ta có: \(\frac{201-x}{99}+\frac{203-x}{97}=\frac{205-x}{95}+3=0\)
\(\Leftrightarrow\frac{201-x}{99}+\frac{203-x}{97}-\frac{205-x}{95}-3=0\)
\(\Leftrightarrow\left(\frac{201-x}{99}+1\right)+\left(\frac{203-x}{97}+1\right)+\left(\frac{205-x}{95}+1\right)=0\)
\(\Leftrightarrow\frac{201-x+99}{99}+\frac{203-x+97}{97}+\frac{205-x+95}{95}=0\)
\(\Leftrightarrow\frac{300-x}{99}+\frac{300-x}{97}+\frac{300-x}{95}=0\)
\(\Leftrightarrow\left(300-x\right)\left(\frac{1}{99}+\frac{1}{97}+\frac{1}{95}\right)=0\)
Vì \(\frac{1}{99}+\frac{1}{97}+\frac{1}{95}\ne0\)
nên 300-x=0
\(\Leftrightarrow x=300\)
Vậy: x=300
c) Ta có: \(x^3+x^2+x+1=0\)
\(\Leftrightarrow x^2\left(x+1\right)+\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^2+1\right)=0\)(1)
Ta có: \(x^2\ge0\forall x\)
\(\Rightarrow x^2+1\ge1\ne0\forall x\)(2)
Từ (1) và (2) suy ra x+1=0
hay x=-1
Vậy: x=-1
d) Ta có: \(\left(x-1\right)x\left(x+1\right)\left(x+2\right)=24\)
\(\Leftrightarrow\left(x^2+x\right)\left(x^2+x-2\right)=24\)
Đặt \(x^2+x-1=t\)
\(\Leftrightarrow\left(t+1\right)\left(t-1\right)=24\)
\(\Leftrightarrow t^2-1-24=0\)
\(\Leftrightarrow t^2-25=0\)
\(\Leftrightarrow\left(t-5\right)\left(t+5\right)=0\)
\(\Leftrightarrow\left(x^2+x-1-5\right)\left(x^2+x-1+5\right)=0\)
\(\Leftrightarrow\left(x^2+x-6\right)\left(x^2+x+4\right)=0\)
\(\Leftrightarrow\left(x^2+3x-2x-6\right)\left(x^2+2\cdot x\cdot\frac{1}{2}+\frac{1}{4}+\frac{15}{4}\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x-2\right)\left[\left(x+\frac{1}{2}\right)^2+\frac{15}{4}=0\right]\)(3)
Ta có: \(\left(x+\frac{1}{2}\right)^2\ge0\forall x\)
\(\Rightarrow\left(x+\frac{1}{2}\right)^2+\frac{15}{4}\ge\frac{15}{4}\ne0\forall x\)(4)
Từ (3) và (4) suy ra
\(\left[{}\begin{matrix}x+3=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=2\end{matrix}\right.\)
Vậy: \(x\in\left\{-3;2\right\}\)
e) Ta có: \(\left(5x-3\right)-\left(4x-7\right)=0\)
\(\Leftrightarrow5x-3-4x+7=0\)
\(\Leftrightarrow x+4=0\)
\(\Leftrightarrow x=-4\)
Vậy: x=-4
f) Ta có: \(3x^2+2x-1=0\)
\(\Leftrightarrow3x^2+3x-x-1=0\)
\(\Leftrightarrow3x\left(x+1\right)-\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(3x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\3x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\3x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\frac{1}{3}\end{matrix}\right.\)
Vậy: \(x\in\left\{-1;\frac{1}{3}\right\}\)
g) Ta có: \(x^2+6x-16=0\)
\(\Leftrightarrow x^2-2x+8x-16=0\)
\(\Leftrightarrow x\left(x-2\right)+8\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+8\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+8=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-8\end{matrix}\right.\)
Vậy: \(x\in\left\{2;-8\right\}\)
h) Ta có: \(x^2+3x-10=0\)
\(\Leftrightarrow x^2+5x-2x-10=0\)
\(\Leftrightarrow x\left(x+5\right)-2\left(x+5\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+5=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)
Vậy: \(x\in\left\{-5;2\right\}\)
i) Ta có: \(x^2+x-2=0\)
\(\Leftrightarrow x^2-x+2x-2=0\)
\(\Leftrightarrow x\left(x-1\right)+2\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)
Vậy: \(x\in\left\{1;-2\right\}\)
k) Ta có: \(3x^2+7x+2=0\)
\(\Leftrightarrow3x^2+6x+x+2=0\)
\(\Leftrightarrow3x\left(x+2\right)+\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(3x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\3x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\3x=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\frac{-1}{3}\end{matrix}\right.\)
Vậy: \(x\in\left\{-2;\frac{-1}{3}\right\}\)
l) Ta có: \(4x^2-12x+5=0\)
\(\Leftrightarrow4x^2-2x-10x+5=0\)
\(\Leftrightarrow2x\left(2x-1\right)-5\left(2x-1\right)=0\)
\(\Leftrightarrow\left(2x-1\right)\left(2x-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\2x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=1\\2x=5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{2}\\x=\frac{5}{2}\end{matrix}\right.\)
Vậy: \(x\in\left\{\frac{1}{2};\frac{5}{2}\right\}\)
a/ \(x^2-5x+5y-y^2=\left(x^2-y^2\right)-\left(5x-5y\right)=\left(x-y\right)\left(x+y\right)-5\left(x-y\right)=\left(x-y\right)\left(x+y-5\right)\)
b/ \(3x^2-6xy+3y^2-12z^2=3\left(x^2-2xy+y^2-4z^2\right)=3\left[\left(x^2-2xy+y^2\right)-\left(2x\right)^2\right]=3\left[\left(x-y\right)^2-\left(2x\right)^2\right]=3\left(x-y-2x\right)\left(x-y+2x\right)=3\left(-x-y\right)\left(3x-y\right)\)
c/ \(x^2-2xy+y^2-xz+yz=\left(x^2-2xy+y^2\right)-\left(xz-yz\right)=\left(x-y\right)^2-z\left(x-y\right)=\left(x-y\right)\left(x-y-z\right)\)
d/ \(x^2-x+2y-4y^2=\left(x^2-4y^2\right)-\left(x+2y\right)=\left(x+2y\right)\left(x-2y\right)-\left(x+2y\right)=\left(x+2y\right)\left(x-2y-1\right)\)
e/ \(x^6-y^6=\left(x^3\right)^2-\left(y^3\right)^2=\left(x^3-y^3\right)\left(x^3+y^3\right)\)
a) x2 - 5x + 5y - y2
= ( x2 - y2 ) - ( 5x - 5y )
= ( x - y )( x + y ) - 5( x - y )
= ( x - y )( x + y - 5 )
b) 3x2 - 6xy + 3y2 - 12z2
= 3( x2 - 2xy + y2 - 4z2 )
= 3[( x2 - 2xy + y2 ) - 4z2 ]
= 3[( x - y )2 - 4z2 ]
= 3( x - y - 2z )( x - y + 2z )
c) x2 - 2xy + y2 - xz - yz
= ( x2 - 2xy + y2 ) - ( xz - yz )
= ( x - y )2 - z( x - y )
= ( x - y )( x - y - z )
d) x2 - x + 2y - 4y2
= ( x2 - 4y2 ) - ( x - 2y )
= ( x - 2y )( x + 2y ) - ( x - 2y )
= ( x - 2y )(x + 2y - 1 )
e) x6 - y6
= ( x3 )2 - ( y3 )2
= ( x3 - y3 )( x3 + y3 )
= ( x - y )( x2 + xy + y2 )( x + y )( x2 - xy + y2 )
Chúc bạn học tốt 
làm nốt
d) (2x-1)(3x+2)(3-x)
=(6x2+x-2)(3-x)
=-6x3+17x2+5x-6
e) (x+3)(x2+3x-5)
=x3+6x2+4x-15
f) (xy-2)(x3-2x-6)
=x4y-2x3-2x2y-6xy+4x+12
g) (5x3-x2+2x-3)(4x2-x+2)
=20x5-9x4+19x3-16x2+7x-6
Bài 1:
a) (x-2)(x2+3x+4)
=x(5x+4)-2(5x+4)
= 5x2+4x-10x-8
=5x2-6x-8
a, 3x - 3y = 3( x- y )
b, x2 - x =x(x - 1)
c, 3(x - y) - 5x(y - x)
= 3(x - y) + 5x(x - y)
= ( x - y)(3 + 5x)
d, x(y - 1) - y(y - 1)
= (x - y)(y - 1)
e, 10x(x - y)-8y( y - x)
= 10x(x - y) + 8y(x - y)
= (10y + 8x)(x - y)
f, 2x2 +5x3 +xy
= x(2x + 5x2 + y)
g, 14x2y - 21xy2 +28x2y2
= 7xy(2x - 3y + 4xy)
h, x2 - 3x + 2
= x2 - x - 2x + 2
= x(x - 1)- 2(x - 1)
= (x - 2)(x - 1)
i, x2 - x - 6
x2 + 2x - 3x - 6
x(x + 2) - 3(x + 2)
(x + 2)(x - 3)
k, x2 + 5x+6
= x2 - x + 6x + 6
=x(x - 1) + 6(x + 1)
= x(x - 1) - 6(x - 1)
= (x - 6)(x - 1)
l,x2 - 4x + 3
= x2 - x - 3x + 3
= x(x - 1) - 3(x - 1)
= (x - 3)(x - 1)
m, x2 + 5x +4
= x2 + x + 4x + 4
= x(x + 1) + 4(x + 1)
= (x + 4)(x + 1)
Hướng dẫn:
a, b, c, d, e, f, g: Phương pháp phân phối đưa thừa số chung ra ngoài
h, i, k, l, m : Tách hạng tử rồi nhóm
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