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a, (1-x)(5x+3)= (3x-8)(1-x)
<=> (1-x) (5x+3) - (3x-8)(1-x) =0 <=> (1-x) (2x+11) = 0
\(\Leftrightarrow\left[{}\begin{matrix}1-x=0\\2x+11=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{11}{2}\end{matrix}\right.\)
Vậy.........
b, (x-3)(x+4)-2(3x-2)=(x-4)^2
<=> 3x = 24<=> x=8
Vậy .......
c,x^2+ x^3+x+1=0
<=> x^2 (x+1) +(x+1) =0 <=> (x^2 +1)(x+1) =0
<=> x+1 =0 => x=-1
Vậy.......
d, \(\dfrac{x-3}{x+3}-\dfrac{2}{x-3}=\dfrac{3x+1}{9-x^2}\)
\(\Leftrightarrow x^2-6x+9-2x-6=-3x-1\)
\(\Leftrightarrow x^2-5x+4=0\Leftrightarrow\left(x+4\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+4=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=1\end{matrix}\right.\)
Vậy...........
a: =>4(2x-1)-12x=3(x+3)+24
=>8x-4-12x=3x+9+24
=>-4x-4=3x+33
=>-7x=37
=>x=-37/7
b: =>(x-2)(x+2+x-9)=0
=>(2x-7)(x-2)=0
=>x=2 hoặc x=7/2
c: =>(x-1)(x+3)-x+3=3x+3
=>x^2+2x-3-x+3=3x+3
=>x^2+x-3x-3=0
=>x^2-2x-3=0
=>(x-3)(x+1)=0
=>x=-1
\(\left(x+1\right)^2-3\left(x+1\right)=\left(x+1\right)\left(x+1-3\right)=\left(x+1\right)\left(x-2\right)\)
\(2x\left(x-2\right)-\left(x-2\right)^2=\left(x-2\right)\left[2x-\left(x-2\right)\right]=\left(x-2\right)\left(2x-x+2\right)=\left(x-2\right)\left(x+2\right)\)
\(4x^2-20xy+25y^2=\left(2x\right)^2-2.2x.5y+\left(5y\right)^2=\left(2x-5y\right)^2\)
\(x^2+3x-x-3=x\left(x+3\right)-\left(x+3\right)=\left(x-1\right)\left(x+3\right)\)
\(x^2-xy+x-y=x\left(x-y\right)+\left(x-y\right)=\left(x-y\right)\left(x+1\right)\)
\(2y\left(x+2\right)-3x-6=2y\left(x+2\right)-3\left(x+2\right)=\left(x+2\right)\left(2y-3\right)\)
a)(3x-1)(4x-8)=0
⇔3x-1=0 hoặc 4x-8=0
1.3x-1=0⇔3x=1⇔x=1/3
2.4x-8=0⇔4x=8⇔x=2
phương trình có 2 nghiệm:x=1/3 và x=2
b)(x-2)(1-3x)=0
⇔x-2=0 hoặc 1-3x=0
1.x-2=0⇔x=2
2.1-3x=0⇔-3x=1⇔x=-1/3
phương trình có 2 nghiệm:x=2 và x=-1/3
c)(x-3)(x+4)-(x-3)(2x-1)=0
⇔(x+4)(2x-1)=0
⇔x+4=0 hoặc 2x-1=0
1.x+4=0⇔x=-4
2.2x-1=0⇔2x=1⇔x=1/2
phương trình có hai nghiệm:x=-4 và x=1/2
d)(x+1)(x+2)=2x(x+2)
⇔(x+1)(x+2)-2x(x+2)=0
⇔2x(x+1)=0
⇔2x=0 hoặc x+1=0
1.2x=0⇔x=0
2.x+1=0⇔x=-1
phương trình có 2 nghiệm:x=0 và x=-1
a) \(2\left(x-2\right)+x-2=3\left(x-2\right)\)
\(\Leftrightarrow\left(2+1\right)\left(x-2\right)=3\left(x-2\right)\)
Vì phương trình trên luôn đúng với mọi x nên có vô số nghiệm
B) \(4\left(1-x\right)+3x=1-x\)
\(4-4x+3x=1-x\Leftrightarrow4-x=1-x\)(vô nghiệm)
\(A=\left(x-2\right)^2-\left(2x+1\right)^2=x^2-4x+4-4x^2-4x-1=-3x^2+3=-3\left(x^2-1\right)\)
\(=-3\left(x-1\right)\left(x+1\right)\)
\(B=\left(x-2y\right)^2-\left(x-2y\right)\left(x+2y\right)=\left(x-2y\right)\left(x-2y-x-2y\right)=-4y\left(x-2y\right)\)
\(C=\left(x+1\right)^3-\left(x-2\right)^3=\left(x^3+3x^2+3x+1\right)-\left(x^3-6x^2+12x-8\right)\)
\(=x^3+3x^2+3x+1-x^3+6x^2-12x+8=9x^2-9x+9=9\left(x^2-x+1\right)\)
\(D=\left(x-1\right)^2-2\left(x-1\right)\left(x+1\right)+\left(x+1\right)^2=\left(x-1-x-1\right)^2=-2^2=4\)
\(E=\left(x+2y\right)^2+2\left(x+2y\right)\left(x-2y\right)+2y-x=x^2+4xy+4y^2+2\left(x^2-4y^2\right)+2y-x\)
\(=x^2+4xy+4y^2+2x^2-8y^2+2y-x=3x^2-4y^2+4xy+2y-x\)
\(G=\left(2x+1\right)^3-\left(2x-1\right)=8x^3+12x^2+6x+1-2x+1=8x^3+12x^2+4x+2\)
\(=2\left(4x^3+6x^2+2x+1\right)=2\left(4x\left(x+1\right)^2+1\right)\)