Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a: \(\Leftrightarrow4x^2-20x+25-4x^2+9=5\)
=>-20x+34=5
=>-20x=-29
hay x=29/20
b: \(\Leftrightarrow x^3+9x^2+27x+27-x\left(x^2-16\right)=27\)
\(\Leftrightarrow x^3+9x^2+27x+27-x^3+16x-27=0\)
\(\Leftrightarrow9x^2+43x=0\)
=>x(9x+43)=0
=>x=0 hoặc x=-43/9
c: \(\Leftrightarrow9x^2-12x+4-9x^2-30x-25=4\)
=>-42x-21=4
=>-42x=25
hay x=-25/42
d: \(\Leftrightarrow\left(3x+5+x-3\right)\left(3x+5-x+3\right)=0\)
=>(4x+2)(2x+8)=0
=>x=-1/2 hoặc x=-4
(x+2)(x+3)-(x-2)(x+5)=0
=> x2+5x+6-x2-3x+10=0
=>2x+16=0
=>2x=-16
=>x=-8
\(1,A=\left(3x+7\right)\left(2x+3\right)-\left(2x+3\right)-\left(3x-5\right)\left(2x+11\right)\\ =6x^2+23x+21-2x-3-6x^2-23x+55\\ =73-2x\left(đề.sai\right)\\ B=x^4+x^3-x^2-2x^2-2x+2-x^4-x^3+3x^2+2x\\ =2\\ 2,\\ a,\Leftrightarrow30x^2+18x+3x-30x^2=7\\ \Leftrightarrow21x=7\Leftrightarrow x=\dfrac{1}{3}\\ b,\Leftrightarrow-63x^2+78x-15+63x^2+x-20=44\\ \Leftrightarrow79x=79\Leftrightarrow x=1\\ c,\Leftrightarrow\left(x+5\right)\left(x^2+3x+2\right)-x^3-8x^2=27\\ \Leftrightarrow x^3+3x^2+2x+5x^2+15x+10-x^3-8x^2=27\\ \Leftrightarrow17x=17\Leftrightarrow x=1\)
\(d,\Leftrightarrow7x-2x^2-3+x^2+x-6=-x^2-x+2\\ \Leftrightarrow9x=11\Leftrightarrow x=\dfrac{11}{9}\)
a: \(\Leftrightarrow4x^2-20x+25-4x^2+9=5\)
=>-20x+34=5
=>-20x=-29
hay x=29/20
b: \(\Leftrightarrow x^3+9x^2+27x+27-x\left(x^2-16\right)=27\)
\(\Leftrightarrow x^3+9x^2+27x+27-x^3+16x-27=0\)
\(\Leftrightarrow9x^2+43x=0\)
=>x(9x+43)=0
=>x=0 hoặc x=-43/9
c: \(\Leftrightarrow9x^2-12x+4-9x^2-30x-25=4\)
=>-42x-21=4
=>-42x=25
hay x=-25/42
d: \(\Leftrightarrow\left(3x+5+x-3\right)\left(3x+5-x+3\right)=0\)
=>(4x+2)(2x+8)=0
=>x=-1/2 hoặc x=-4
\(a,3\left(2x-3\right)+2\left(2-x\right)=-3\\ \Leftrightarrow6x-9+4-2x=-3\\ \Leftrightarrow4x=2\\ \Leftrightarrow x=\dfrac{1}{2}\\ b,x\left(5-2x\right)+2x\left(x-1\right)=13\\ \Leftrightarrow5x-2x^2+2x^2-2x=13\\ \Leftrightarrow3x=13\\ \Leftrightarrow x=\dfrac{13}{3}\\ c,5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\\ \Leftrightarrow5x^2-5x-5x^2-3x+14=6\\ \Leftrightarrow-8x=-8\\ \Leftrightarrow x=1\\ d,3x\left(2x+3\right)-\left(2x+5\right)\left(3x-2\right)=8\\ \Leftrightarrow6x^2+9x-6x^2-11x+10=8\\ \Leftrightarrow-2x=-2\\ \Leftrightarrow x=1\)
\(e,2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\\ \Leftrightarrow10x-16-12x+15=12x-16+11\\ \Leftrightarrow-14x=-4\\ \Leftrightarrow x=\dfrac{2}{7}\\ f,2x\left(6x-2x^2\right)+3x^2\left(x-4\right)=8\\ \Leftrightarrow12x^2-4x^3+3x^3-12x^2=8\\ \Leftrightarrow-x^3-8=0\\ \Leftrightarrow-\left(x^3+8\right)=0\\ \Leftrightarrow-\left(x+2\right)\left(x^2-2x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-2\\x\in\varnothing\left(x^2-2x+4=\left(x-1\right)^2+3>0\right)\end{matrix}\right.\)
Bài 4:
a: Ta có: \(3\left(2x-3\right)-2\left(x-2\right)=-3\)
\(\Leftrightarrow6x-9-2x+4=-3\)
\(\Leftrightarrow4x=2\)
hay \(x=\dfrac{1}{2}\)
b: Ta có: \(x\left(5-2x\right)+2x\left(x-1\right)=13\)
\(\Leftrightarrow5x-2x^2+2x^2-2x=13\)
\(\Leftrightarrow3x=13\)
hay \(x=\dfrac{13}{3}\)
c: Ta có: \(5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\)
\(\Leftrightarrow5x^2-5x-5x^2+7x-10x+14=6\)
\(\Leftrightarrow-8x=-8\)
hay x=1
a: \(\Leftrightarrow\left(x+2\right)\left(x+2-2x+10\right)=0\)
\(\Leftrightarrow x\in\left\{-2;12\right\}\)
a) (2x-5)2-(2x-3)(2x+3)=5
(2x-5)2-[(2x)2-32 ]=5
(2x-5)2-(2x)2+9=5
[(2x-5)-(2x)][(2x-5)+2x]=5-9
(2x-5-2x)(2x-5+2x)=-4
-5(4x-5)=-4
4x-5=-4:-5
4x-5=\(\dfrac{4}{5}\)
4x=\(\dfrac{4}{5}\)+5
4x=\(\dfrac{29}{5}\)
x=\(\dfrac{29}{5}\):4
x=\(\dfrac{29}{20}\)
vậy x=\(\dfrac{29}{20}\)