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Bài 1:
a) \(3x^2-2x(5+1,5x)+10=3x^2-(10x+3x^2)+10\)
\(=10-10x=10(1-x)\)
b) \(7x(4y-x)+4y(y-7x)-2(2y^2-3,5x)\)
\(=28xy-7x^2+(4y^2-28xy)-(4y^2-7x)\)
\(=-7x^2+7x=7x(1-x)\)
c)
\(\left\{2x-3(x-1)-5[x-4(3-2x)+10]\right\}.(-2x)\)
\(\left\{2x-(3x-3)-5[x-(12-8x)+10]\right\}(-2x)\)
\(=\left\{3-x-5[9x-2]\right\}(-2x)\)
\(=\left\{3-x-45x+10\right\}(-2x)=(13-46x)(-2x)=2x(46x-13)\)
Bài 2:
a) \(3(2x-1)-5(x-3)+6(3x-4)=24\)
\(\Leftrightarrow (6x-3)-(5x-15)+(18x-24)=24\)
\(\Leftrightarrow 19x-12=24\Rightarrow 19x=36\Rightarrow x=\frac{36}{19}\)
b)
\(\Leftrightarrow 2x^2+3(x^2-1)-5x(x+1)=0\)
\(\Leftrightarrow 2x^2+3x^2-3-5x^2-5x=0\)
\(\Leftrightarrow -5x-3=0\Rightarrow x=-\frac{3}{5}\)
\(2x^2+3(x^2-1)=5x(x+1)\)
d)\(x^2-y^2+2x-4y-10=0\)
\(\Leftrightarrow\left(x^2+2x+1\right)-\left(y^2+4y+4\right)=7\)
\(\Leftrightarrow\left(x+1\right)^2-\left(y+2\right)^2=7\)
\(\Leftrightarrow\left(x-y-1\right)\left(x+y+3\right)=7\)
Mà x,y nguyên dương\(\Rightarrow x-y-1< x+y+3\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-y-1=1\\x+y+3=7\end{matrix}\right.\\\left\{{}\begin{matrix}x-y-1=-7\\x+y+3=-1\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=1\end{matrix}\right.\)
Mạn phép ko chép lại đề :
b) \(8\left(x+\dfrac{1}{x}\right)^2+4\left(x^2+\dfrac{1}{x^2}\right)\left(x^2+\dfrac{1}{x^2}-x^2-2-\dfrac{1}{x^2}\right)=\left(x+4\right)^2\)
⇔ \(8\left(x+\dfrac{1}{x}\right)^2-8\left(x^2+\dfrac{1}{x^2}\right)=\left(x+4\right)^2\) ( x # 0)
⇔ \(8\left(x^2+2+\dfrac{1}{x^2}-x^2-\dfrac{1}{x^2}\right)=\left(x+4\right)^2\)
⇔ ( x + 4)2 = 16
⇔ x2 + 8x + 16 = 16
⇔ x( x + 8) = 0
⇔ x = 0 ( KTM) hoặc : x = - 8 ( TM)
KL.....
a: \(A=2x^2-2xy-y^2+2xy=2x^2-y^2\)
\(=2\cdot\dfrac{4}{9}-\dfrac{1}{9}=\dfrac{7}{9}\)
b: \(B=5x^2-20xy-4y^2+20xy=5x^2-4y^2\)
\(=5\cdot\dfrac{1}{25}-4\cdot\dfrac{1}{4}\)
=1/5-1=-4/5
c \(C=x^3+6x^2+12x+8=\left(x+2\right)^3=\left(-9\right)^3=-729\)
d: \(D=20x^3-10x^2+5x-20x^2+10x+4\)
\(=20x^3-30x^2+15x+4\)
\(=20\cdot5^3-30\cdot5^2+15\cdot2+4=1784\)
b: Đặt \(x^2-6x-2=a\)
Theo đề, ta có: \(a+\dfrac{14}{a+9}=0\)
=>(a+2)(a+7)=0
\(\Leftrightarrow\left(x^2-6x\right)\left(x^2-6x+5\right)=0\)
=>x(x-6)(x-1)(x-5)=0
hay \(x\in\left\{0;1;6;5\right\}\)
c: \(\Leftrightarrow\dfrac{-8x^2}{3\left(2x-1\right)\left(2x+1\right)}=\dfrac{2x}{3\left(2x-1\right)}-\dfrac{8x+1}{4\left(2x+1\right)}\)
\(\Leftrightarrow-32x^2=8x\left(2x+1\right)-3\left(8x+1\right)\left(2x-1\right)\)
\(\Leftrightarrow-32x^2=16x^2+8x-3\left(16x^2-8x+2x-1\right)\)
\(\Leftrightarrow-48x^2=8x-48x^2+18x+3\)
=>26x=-3
hay x=-3/26
Bài 1 :
a ) \(2x\left(x+1\right)+2\left(x+1\right)=\left(x+1\right)\left(2x+2\right)=2\left(x+1\right)^2\)
b ) \(y^2\left(x^2+y\right)-zx^2-zy=y^2\left(x^2+y\right)-z\left(x^2+y\right)=\left(x^2+y\right)\left(y^2-z\right)\)
c ) \(4x\left(x-2y\right)+8y\left(2y-x\right)=4x\left(x-2y\right)-8y\left(x-2y\right)=4\left(x-2y\right)^2\)
d ) \(3x\left(x+1\right)^2-5x^2\left(x+1\right)+7\left(x+1\right)=\left(x+1\right)\left(3x^2+3x-5x^2+7\right)=\left(x+1\right)\left(3x-2x^2+7\right)\)
e ) \(x^2-6xy+9y^2=\left(x-3x\right)^2\)
Bài 1 :
f ) \(x^3+6x^2y+12xy^2+8y^3=\left(x+2y\right)^3\)
g ) \(x^3-64=\left(x-4\right)\left(x^2+4x+16\right)\)
h ) \(125x^3+y^6=\left(5x+y^2\right)\left(25x^2-5xy^2+y^4\right)\)
B = (x-1)(2x+1) - (x2-2x-1)
B = 2x2+x-2x-1-x2-2x-1 = x2-3x-2
B = x2+x-4x-2 = x(x+1) - 4(x+1)
B = (x+1)(x-4)
\(A=2x\left(x-2\right)-x\left(2x-3\right)\\ =2x^2-4x-2x^2+3x\\ =-x\\ B=\left(x-1\right)\left(2x+1\right)-\left(x^2-2x-1\right)\\ =x\left(2x+1\right)-\left(2x+1\right)-x^2+2x+1\\ =2x^2+x-2x-1-x^2+2x+1\\ =x^2+x\\ C=\left(x+y\right)\left(x^2-xy+y^2\right)-x^3\\ =x\left(x^2-xy+y^2\right)+y\left(x^2-xy+y^2\right)-x^3\\ =x^3-x^2y+xy^2+x^2y-xy^2+y^3-x^3\\ =y^3\)
\(D=\left(12x-3\right)\left(x+4\right)-x\left(2x+7\right)\\ =x\left(12x-3\right)+4\left(12x-3\right)-2x^2-7x\\ =12x^2-3x+48x-12-2x^2-7x\\ =10x^2+38x-12\\ E=\left(2x+y\right)\left(4x^2-2xy+y^2\right)\\ =2x\left(4x^2-2xy+y^2\right)+y\left(4x^2-2xy+y^2\right)\\ =8x^3-4x^2y+2xy^2+4x^2y-2xy^2+y^3\\ =8x^3+y^3\)
Bài 2:
a, \(x^2-6x+10=x^2-6x+9+1\)
\(=\left(x-3\right)^2+1\ge1>0\)
\(\Rightarrowđpcm\)
b, \(x^2-4xy+4y^2+1=\left(x-2y\right)^2+1>0\)
\(\Rightarrowđpcm\)
c, \(x^2-4x+7=x^2-4x+4+3\)
\(=\left(x-2\right)^2+3\ge3\)
\(\Rightarrowđpcm\)
d, \(x^2+y^2-2x+4y+5\)
\(=x^2-2x+1+y^2+4y+4\)
\(=\left(x-1\right)^2+\left(y+2\right)^2\ge0\)
\(\Rightarrowđpcm\)
Ép người quá đáng >.<
Bài 1:
a, \(-\left(2x^2+2x+1\right)\left(2x^2-2x+1\right)+\left(2x^2+1\right)\)
\(=-\left(4x^4-4x^3+2x^2+4x^3-4x^2+2x+2x^2-2x+1\right)+2x^2+1\)
\(=-\left(4x^4+1\right)+2x^2+1=-4x^4+2x^2\)
b, \(\left(x^2+x+2\right)^2+\left(x-1\right)^2-2\left(x^2+x+2\right)\left(x-1\right)\)
\(=\left(x^2+x+2-x+1\right)^2=\left(x^2+3\right)^2\)
d, \(-125x^3+225x^2-135x+27\)
\(=-\left(125x^3-225x^2+135x-27\right)\)
\(=-\left(125x^3-75x^2-150x^2+90x+45x-27\right)\)
\(=-\left[25x^2\left(5x-3\right)-30x\left(5x-3\right)+9\left(5x-3\right)\right]\)
\(=-\left[\left(5x-3\right)\left(25x^2-15x-15x+9\right)\right]\)
\(=-\left(5x-3\right)^3\)