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13 tháng 8 2016
a,(2x-y)2+(2x+y)2=(2x2-2*2xy+y2)+(2x2+2*2xy+y2)
=2x2-4xy+y2+2x2+4xy+y2
=4x2+2y2
11 tháng 12 2022
a: \(=\dfrac{6x^2-3x+4x^2+2x}{\left(2x-1\right)\left(2x+1\right)}\cdot\dfrac{\left(2x-1\right)^2}{2x\left(4x+5\right)}\)
\(=\dfrac{10x^2+x}{\left(2x+1\right)}\cdot\dfrac{2x-1}{2x\left(4x+5\right)}\)
\(=\dfrac{\left(10x^2+x\right)\left(2x-1\right)}{2x\cdot\left(2x+1\right)\left(4x+5\right)}\)
b: \(=\left(\dfrac{x}{\left(5x-1\right)\left(5x+1\right)}\cdot\dfrac{x\left(5x+1\right)}{5x}\right)\cdot\dfrac{x\left(5x+1\right)}{5x-1}+\dfrac{x}{5x-1}\)
\(=\dfrac{x}{5\left(5x-1\right)}\cdot\dfrac{x\left(5x+1\right)}{5x-1}+\dfrac{x}{5x-1}\)
\(=\dfrac{x^2\left(5x+1\right)+5x\left(5x-1\right)}{5\left(5x-1\right)^2}\)
\(=\dfrac{5x^3+x^2+25x^2-5x}{5\left(5x-1\right)^2}=\dfrac{5x^3+26x^2-5x}{5\left(5x-1\right)^2}\)
c: \(=\dfrac{x+1}{x-2}+\dfrac{1-3x}{x\left(x^2+1\right)}\cdot\dfrac{x^2+1}{x-1}\)
\(=\dfrac{x+1}{x-2}+\dfrac{1-3x}{x\left(x-1\right)}\)
\(=\dfrac{x^3-x+\left(1-3x\right)\left(x-2\right)}{x\left(x-1\right)\left(x-2\right)}\)
\(=\dfrac{x^3-x+x-2-3x^2+6x}{x\left(x-1\right)\left(x-2\right)}=\dfrac{x^3-3x^2+6x-2}{x\left(x-1\right)\left(x-2\right)}\)
PN
17 tháng 8 2020
a, \(12-2\left(1-x\right)^2=\left(3x-2\right)\left(2x-3\right)\)
\(< =>12-2\left(1-2x+x^2\right)=6x^2-9x-4x+6\)
\(< =>12-2+4x-2x^2=6x^2-13x+6\)
\(< =>10+4x-2x^2-6x^2+13x-6=0\)
\(< =>-8x^2+17x+4=0< =>\orbr{\begin{cases}x=\frac{17-\sqrt{417}}{16}\\x=\frac{17+\sqrt{417}}{16}\end{cases}}\)
b, \(10x+3-5x=4x+12< =>5x+3-4x-12=0\)
\(< =>x-9=0< =>x=9\)
c, \(11x+42-2x=100-9x-22< =>9x+42-100+9x+22=0\)
\(< =>18x+64-100=0< =>18x-36=0< =>x=\frac{36}{18}=2\)
d, \(2x-\left(3-5x\right)=4\left(x+3\right)< =>2x-3+5x=4x+12\)
\(< =>7x-3-4x-12=0< =>3x-15=0< =>x=\frac{15}{3}=5\)
e, \(2\left(x-3\right)+5x\left(x-1\right)=5x^2< =>2x-6+5x^2-5=5x^2\)
\(< =>2x-11+5x^2-5x^2=0< =>2x-11=0< =>x=\frac{11}{2}\)
f, \(-6\left(1,5-2x\right)=3\left(-15+2x\right)< =>-6\left(\frac{3}{2}-2x\right)=3\left(2x-15\right)\)
\(< =>-9+12x-6x+45=0< =>6x+36=0< =>x=-6\)
g, \(14x-\left(2x+7\right)=3x+12x-13< =>14x-2x-7=15x-13\)
\(< =>12x-7-15x+13=0< =>-3x+6=0< =>x=-2\)
h, \(\left(x-4\right)\left(x+4\right)-2\left(3x-2\right)=\left(x-4\right)^2\)
\(< =>x^2-16-6x+4=x^2-8x+16\)
\(< =>x^2-6x-12-x^2+8x-16=0\)
\(< =>2x-28=0< =>x=\frac{28}{2}=14\)
q, \(4\left(x-2\right)-\left(x-3\right)\left(2x-5\right)=?\)thiếu đề
13 tháng 7 2018
Mình giải từ cuối lên , mình giải dần -)
n, <=> x(2x-1)-3(2x-1)=0
<=> (x-3)(2x-1)=0
<=> x= 3 hoặc x= 1/2
m, <=> (x+2)(x2-3x+5)-x2(x+2)=0
<=> (x+2)(x2-3x+5-x2)=0
<=> (x+2)(5-3x)=0
=> x= -2 hoặc5/3
14 tháng 8 2020
a) Ta có: \(\left(x^2-1\right)\left(x^2+2x\right)\)
\(=x^4+2x^3-x^2-2x\)
b) Ta có: \(\left(2x-1\right)\left(3x+2\right)\left(3-x\right)\)
\(=\left(6x^2+4x-3x-2\right)\left(3-x\right)\)
\(=\left(6x^2+x-2\right)\left(3-x\right)\)
\(=18x^2-6x^3+3x-x^2-6+2x\)
\(=-6x^3+17x^2+5x-6\)
c) Ta có: \(\left(x+3\right)\left(x^2+3x-5\right)\)
\(=x^3+3x^2-5x+3x^2+9x-15\)
\(=x^3+6x^2+4x-15\)
d) Ta có: \(\left(x+1\right)\left(x^2-x+1\right)\)
\(=x^3+1\)
e) Ta có: \(\left(2x^3-3x-1\right)\left(5x+2\right)\)
\(=10x^4+4x^3-15x^2-6x-5x-2\)
\(=10x^4+4x^3-15x^2-11x-2\)
f) Ta có: \(\left(x^2-2x+3\right)\left(x-4\right)\)
\(=x^3-4x^2-2x^2+8x+3x-12\)
\(=x^3-6x^2+11x-12\)
g) Ta có: \(\left(4x-1\right)\left(3x+1\right)-5x\left(x-3\right)-\left(x-4\right)\left(x-3\right)\)
\(=12x^2+4x-3x-1-5x^2+15x-\left(x^2-7x+12\right)\)
\(=7x^2+16x-1-x^2+7x-12\)
\(=6x^2+23x-23\)
h) Ta có: \(\left(5x-2\right)\left(x+1\right)-3x\left(x^2-x-3\right)-2x\left(x-5\right)\left(x-4\right)\)
\(=5x^2+5x-2x-2-3x^3+3x^2+9x-2x\left(x^2-9x+20\right)\)
\(=-3x^3+8x^2+12x-2-2x^3+18x^2-40x\)
\(=-5x^3+26x^2-28x-2\)
11 tháng 5 2020
\(x^3-6x^2+5x+12>0\\ < =>\left(x^3-5x-x+5x\right)+12>0\\ < =>\left[\left(x^3-x\right)-\left(5x-5x\right)\right]+12>0\\ < =>x^2+12>0\\ < =>x^2>-12\\ =>x\in R\\ BPTcóvôsốnghiem\)
A
5 tháng 9 2020
a, \(\left(x+1\right)\left(x-2\right)=x^2-2x+x-2=x^2-x-2\)
b, \(-7x^2\left(3x-4y\right)=-21x^3+28x^2y\)
c, \(\left(x+4\right)\left(x-2\right)-\left(x-3\right)^2=x^2-2x+4x-8-\left(x^2-6x+9\right)\)
\(=x^2+2x-8-x^2+6x-9=8x-17\)
Q
18 tháng 7 2017
a) \(\left(4x-1\right)^2-\left(3x+2\right)\left(3x-2\right)=\left(7x-1\right)\left(x+2\right)+\left(2x+1\right)^2-\left(4x^2+7\right)\)(1)
\(\Leftrightarrow\left(16x^2-8x+1\right)-\left(9x^2-4\right)=\left(7x^2+14x-x-2\right)+\left(4x^2+4x+1\right)-\left(4x^2+7\right)\)
\(\Leftrightarrow16x^2-8x+1-9x^2+4=7x^2+13x-2+4x^2+4x+1-4x^2-7\)
\(\Leftrightarrow7x^2-8x+5=7x^2+17x-8\)
\(\Leftrightarrow7x^2-8x-7x^2-17x=-8-5\)
\(\Leftrightarrow-25x=-13\)
\(\Leftrightarrow x=\dfrac{13}{25}\)
Vậy tập nghiệm phương trình (1) là \(S=\left\{\dfrac{13}{25}\right\}\)
\(A=15x^3-6x^2-3x\)
\(B=2x^3-3x-5x^3-x^2+x^2=-3x^3-3x\)
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