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\(7x\left(2-3x\right)+x^2\left(2x+1\right)-2x^2\left(x-2\right)+2x\left(8x-7\right)=14x-21x^2+2x^3+x^2-2x^3+4^2+16x^2-14x=0\)
\(\frac{3x^4-8x^3-10x^2+8x-5}{3x^2-2x+1}\)
\(=\frac{x^2\left(3x^2-2x+1\right)-2x\left(3x^2-2x+1\right)-5\left(3x^2-2x+1\right)}{3x^2-2x+1}\)
\(=\frac{\left(3x^2-2x+1\right)\cdot\left(x^2-2x-5\right)}{3x^2-2x+1}\)
\(=x^2-2x-5\)
\(\frac{2x^3-9x^2+19x-15}{x^2-3x+5}\)
\(=\frac{2x\left(x^2-3x+5\right)-3\left(x^2-3x+5\right)}{x^2-3x+5}\)
\(=\frac{\left(x^2-3x+5\right)\left(2x-3\right)}{x^2-3x+5}\)
\(=2x-3\)
\(\left(3x-4\right)^2-4\left(x+1\right)^2=0\)
\(\Leftrightarrow\left(3x-4\right)^2-\left(2x+2\right)^2=0\)
\(\Leftrightarrow\left(3x-4-2x-2\right)\left(3x-4+2x+2\right)=0\)
\(\Leftrightarrow\left(x-6\right)\left(5x-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=6\\x=\frac{2}{5}\end{cases}}\) ( thỏa mãn )
Vậy : ...
1/ \(\left(3x-4\right)^2-4\left(x+1\right)^2=0\)
\(\Leftrightarrow9x^2-24x+16-4\left(x^2+2x+1\right)=0\)
\(\Leftrightarrow9x^2-24x+16-4x^2-8x-4=0\)
\(\Leftrightarrow5x^2-32x+12=0\)
\(\Leftrightarrow5x^2-30x-2x+12=0\)
\(\Leftrightarrow5x\left(x-6\right)-2\left(x-6\right)=0\)
\(\Leftrightarrow\left(x-6\right)\left(5x-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-6=0\\5x-2=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=6\\x=\frac{2}{5}\end{cases}}\)
Vậy tập nghiệm của phương trình là : \(S=\left\{6;\frac{2}{5}\right\}\)
2/ \(x^4+2x^3-3x^2-8x-4=0\)
\(\Leftrightarrow x^4+2x^3-3x^2-6x-2x-4=0\)
\(\Leftrightarrow x^3\left(x+2\right)-3x\left(x+2\right)-2\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x^3-3x-2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x^3+2x^2+x-2x^2-4x-2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left[x\left(x^2+2x+1\right)-2\left(x^2+2x+1\right)\right]=0\)
\(\Leftrightarrow\left(x+2\right)\left(x+1\right)^2\left(x-2\right)=0\)
\(\Leftrightarrow\)\(x+2=0\)
hoặc \(x+1=0\)
hoặc \(x-2=0\)
\(\Leftrightarrow\)\(x=2\)
hoặc \(x=-1\)
hoặc \(x=2\)
Vậy tập nghiệm của phương trình là \(S=\left\{2;-2;-1\right\}\)
a/ \(=8x^3+2x^2-8x^3-8x^2-8x^3-2x+3=-8x^3-6x^2-2x+3\)
b/ \(=3x^2+12x-7x+20+2x^3-3x^2-2x^3-5x=20\)
Biểu thức A phụ thuộc vào x còn B thì không.
a) \(\left(2x+3\right)\left(x-4\right)+\left(x+5\right)\left(x-2\right)=\left(3x-5\right)\left(x-4\right)\)
\(\Leftrightarrow2x^2-8x+3x-12+x^2-2x-5x+10=3x^2-12x-5x+20\)
\(\Leftrightarrow2x^2-8x+3x-12+x^2-2x+10=3x^2-12x+20\)
\(\Leftrightarrow3x^2-7x-2=3x^2-12x+20\)
\(\Leftrightarrow-7x+12x=20+2\)
\(\Leftrightarrow5x=22\)
\(\Rightarrow x=\dfrac{22}{5}\)
tick cho mk nha
b) \(\left(8x-3\right)\left(3x+2\right)-\left(4x+7\right)\left(x+4\right)=\left(2x+1\right)\left(5x-1\right)\)
\(\Leftrightarrow24x^2+16x-9x-6-4x^2-23x-28=10x^2+3x-1\)
\(\Leftrightarrow20x^2-16x-34-10x^2-3x+1=0\)
\(\Leftrightarrow10x^2-19x-33=0\)
\(\Delta=\left(-19\right)^2-4.10.\left(-33\right)=1320\)
\(x_1=3;x_2=\dfrac{-11}{10}\)
Tick cho mk nha
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
Ta có:7x(2-3x)+x2(2x+1)-2x2(x-2)+2x(8x-7)
=14x-21x2+2x3+x2-2x3+4x2+16x2-14x
=(14x-14x)+(-21x2+x2+4x2+16x2)+(2x3-2x3)
=0