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\(1.6x\left(x-10\right)-2x+20=0\)
⇔\(6x\left(x-10\right)-2\left(x-10\right)=0\)
⇔ \(2\left(x-10\right)\left(3x-1\right)=0\)
⇔ x = 10 hoặc x = \(\dfrac{1}{3}\)
KL....
\(2.3x^2\left(x-3\right)+3\left(3-x\right)=0\)
⇔ \(3\left(x-3\right)\left(x^2-1\right)=0\)
⇔ \(x=+-1\) hoặc \(x=3\)
KL....
\(3.x^2-8x+16=2\left(x-4\right)\)
⇔ \(\left(x-4\right)^2-2\left(x-4\right)=0\)
⇔ \(\left(x-4\right)\left(x-6\right)=0\)
⇔ \(x=4\) hoặc \(x=6\)
KL.....
\(4.x^2-16+7x\left(x+4\right)=0\)
\(\text{⇔}4\left(x+4\right)\left(2x-1\right)=0\)
⇔ \(x=-4hoacx=\dfrac{1}{2}\)
KL.....
\(5.x^2-13x-14=0\)
⇔ \(x^2+x-14x-14=0\)
\(\text{⇔}\left(x+1\right)\left(x-14\right)=0\)
\(\text{⇔}x=14hoacx=-1\)
KL......
Còn lại tương tự ( dài quá ~ )
1: \(\Leftrightarrow5x^2+4x-1-2x^2+12x-18=3x^2+5x-2-x^2-8x-16+x^2-x\)
\(\Leftrightarrow3x^2+16x-19=3x^2-4x-18\)
=>20x=1
hay x=1/20
2: \(\Leftrightarrow5x^2-20x-41=x^2-10x+25+4x^2+4x+1-\left(x^2-2x\right)+\left(x-1\right)^2\)
\(\Leftrightarrow5x^2-20x-41=4x^2-4x+26+x^2-2x+1\)
\(\Leftrightarrow-20x-41=-6x+27\)
=>-14x=68
hay x=-34/7
Lời giải:
a) ĐKXĐ: $x\neq \pm 1$
\(\frac{x^4-4x^2+3}{x^4+6x^2-7}=\frac{x^2(x^2-1)-3(x^2-1)}{x^2(x^2-1)+7(x^2-1)}=\frac{(x^2-3)(x^2-1)}{(x^2-1)(x^2+7)}=\frac{x^2-3}{x^2+7}\)
b) ĐKXĐ: Với mọi $x\in\mathbb{R}$
\(\frac{x^4+x^3-x-1}{x^4+x^4+2x^2+x+1}=\frac{(x^4-x)+(x^3-1)}{(x^4+x^3+x^2)+(x^2+x+1)}=\frac{x(x^3-1)+(x^3-1)}{x^2(x^2+x+1)+(x^2+x+1)}\)
\(=\frac{(x^3-1)(x+1)}{(x^2+1)(x^2+x+1)}=\frac{(x-1)(x^2+x+1)(x+1)}{(x^2+1)(x^2+x+1)}=\frac{x^2-1}{x^2+1}\)
c) ĐK: $x\neq 1;-2$
\(\frac{x^3+3x^2-4}{x^3-3x+2}=\frac{x^2(x-1)+4(x^2-1)}{x^2(x-1)+x(x-1)-2(x-1)}=\frac{(x-1)(x^2+4x+4)}{(x-1)(x^2+x-2)}\)
\(=\frac{(x-1)(x+2)^2}{(x-1)(x-1)(x+2)}=\frac{x+2}{x-1}\)
d) ĐK: $x^2+3x-1\neq 0$
\(\frac{x^4+6x^3+9x^2-1}{x^4+6x^3+7x^2-6x+1}=\frac{(x^2+3x)^2-1}{(x^2+3x)^2-2x^2-6x+1}\)
\(=\frac{(x^2+3x-1)(x^2+3x+1)}{(x^2+3x)^2-2(x^2+3x)+1}=\frac{(x^2+3x-1)(x^2+3x+1)}{(x^2+3x-1)^2}=\frac{x^2+3x+1}{x^2+3x-1}\)
Bài 1:tìm x ,biết:
a) (2x - 1)(3x + 2) - 6x(x + 1) = 0
\(\Leftrightarrow6x^2+x-2-6x^2-6x=0\)
\(\Leftrightarrow-5x=2\)
\(\Leftrightarrow x=\frac{-2}{5}\)
b) \(\left(4x-1\right)^2-\left(2x+1\right)\left(8x-3\right)=0\)
\(\Leftrightarrow16x^2-8x+1-16x^2-2x+3=0\)
\(\Leftrightarrow-10x=-4\)
\(\Leftrightarrow x=\frac{2}{5}\)
c) \(4x^2-1=2\left(2x+1\right)\)
\(\Leftrightarrow\left(2x+1\right)\left(2x-1\right)-2\left(2x+1\right)=0\)
\(\Leftrightarrow\left(2x+1\right)\left(2x-3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=-\frac{1}{2}\\x=\frac{3}{2}\end{cases}}\)
2a) \(4x^2-9y^2-6y-1=4x^2-\left(3y+1\right)^2\)
\(=\left(2x-3y-1\right)\left(2x+3y+1\right)\)
b) \(4x^2-1-2x\left(2x-1\right)=\left(2x-1\right)\left(2x+1\right)-2x\left(2x-1\right)\)
\(=1.\left(2x-1\right)\)
c) \(x^2-8x-4y^2+16=\left(x-4\right)^2-4y^2\)
\(=\left(x-4-2y\right)\left(x-4+2y\right)\)
d) \(9x^2-12x-y^2+4=\left(3x-2\right)^2-y^2\)
\(=\left(3x-2-y\right)\left(3x-2+y\right)\)
e) \(4x^2+10x-5=4x^2+2.2.\frac{5}{2}x+\frac{25}{4}-\frac{25}{4}-5\)
\(=\left(2x+\frac{5}{2}\right)^2-\frac{45}{4}\)
\(=\left(2x+\frac{5+3\sqrt{5}}{2}\right)\left(2x+\frac{5-3\sqrt{5}}{2}\right)\)
bn xem lại đề giúp mk
làm phép chia