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Bài 2: Tìm x
a)ĐKXĐ: \(x\ne0\)
Ta có: \(\left(4x^4+3x^3\right):\left(-x^3\right)+\left(15x^2+6x\right):3x=0\)
\(\Leftrightarrow\frac{-x^3\left(4x+3\right)}{x^3}+\frac{3x\left(5x+2\right)}{3x}=0\)
\(\Leftrightarrow-4x-3+5x+2=0\)
\(\Leftrightarrow x-1=0\)
hay x=1(nhận)
Vậy: x=1
b) ĐKXĐ: \(x\notin\left\{0;\frac{1}{3}\right\}\)
Ta có: \(\left(x^2-12x\right):2x-\left(3x-1\right)^2:\left(3x-1\right)=0\)
\(\Leftrightarrow\frac{x\left(x-12\right)}{2x}-\frac{\left(3x-1\right)^2}{\left(3x-1\right)}=0\)
\(\Leftrightarrow\frac{x-12}{x}-3x+1=0\)
\(\Leftrightarrow\frac{x-12}{x}=3x-1\)
\(\Leftrightarrow x-12=x\left(3x-1\right)\)
\(\Leftrightarrow3x^2-x+x-12=0\)
\(\Leftrightarrow3x^2-12=0\)
\(\Leftrightarrow3x^2=12\)
\(\Leftrightarrow x^2=4\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\left(nhận\right)\\x=-2\left(nhận\right)\end{matrix}\right.\)
Vậy: \(x\in\left\{2;-2\right\}\)
`1,(4x^3+3x^3):x^3+(15x^2+6x):(-3x)=0`
`<=> 4 + 3 + (-5x) + (-2)=0`
`<=> -5x+5=0`
`<=>-5x=-5`
`<=>x=1`
`2,(25x^2-10x):5x +3(x-2)=4`
`<=> 5x - 2 + 3x-6=4`
`<=> 8x -8=4`
`<=> 8x=12`
`<=>x=12/8`
`<=>x=3/2`
`3,(3x+1)^2-(2x+1/2)^2=0`
`<=> [(3x+1)-(2x+1/2)][(3x+1)+(2x+1/2)]=0`
`<=>( 3x+1-2x-1/2)(3x+1+2x+1/2)=0`
`<=>( x+1/2) (5x+3/2)=0`
`@ TH1`
`x+1/2=0`
`<=>x=0-1/2`
`<=>x=-1/2`
` @TH2`
`5x+3/2=0`
`<=> 5x=-3/2`
`<=>x=-3/2 : 5`
`<=>x=-15/2`
`4, x^2+8x+16=0`
`<=>(x+4)^2=0`
`<=>x+4=0`
`<=>x=-4`
`5, 25-10x+x^2=0`
`<=> (5-x)^2=0`
`<=>5-x=0`
`<=>x=5`
Ta có: \(\dfrac{4x^4+3x^3}{-x^3}+\dfrac{15x^2+6x}{3x}=0\)
\(\Leftrightarrow-4x-3+5x+2=0\)
\(\Leftrightarrow x-1=0\)
hay x=1
Bài 1.
a. (3x - 1)2 - (x + 3)2 = 0
\(\Leftrightarrow\left(3x-1+x+3\right)\left(3x-1-x-3\right)=0\)
\(\Leftrightarrow\left(4x+2\right)\left(2x-4\right)=0\)
\(\Leftrightarrow4x+2=0\) hoặc \(2x-4=0\)
1. \(4x+2=0\Leftrightarrow4x=-2\Leftrightarrow x=-\dfrac{1}{2}\)
2. \(2x-4=0\Leftrightarrow2x=4\Leftrightarrow x=2\)
S=\(\left\{-\dfrac{1}{2};2\right\}\)
b. \(x^3=\dfrac{x}{49}\)
\(\Leftrightarrow49x^3=x\)
\(\Leftrightarrow49x^3-x=0\)
\(\Leftrightarrow x\left(49x^2-1\right)=0\)
\(\Leftrightarrow x\left(7x+1\right)\left(7x-1\right)=0\)
\(\Leftrightarrow x=0\) hoặc \(7x+1=0\) hoặc \(7x-1=0\)
1. x=0
2. \(7x+1=0\Leftrightarrow7x=-1\Leftrightarrow x=-\dfrac{1}{7}\)
3. \(7x-1=0\Leftrightarrow7x=1\Leftrightarrow x=\dfrac{1}{7}\)
a) Ta có: \(7x^2-28=0\)
\(\Leftrightarrow7\left(x^2-4\right)=0\)
\(\Leftrightarrow7\left(x-2\right)\left(x+2\right)=0\)
mà 7>0
nên (x-2)(x+2)=0
hay \(\left[{}\begin{matrix}x-2=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)
Vậy: \(x\in\left\{2;-2\right\}\)
b) Ta có: \(\dfrac{2}{3}x\left(x^2-4\right)=0\)
\(\Leftrightarrow\dfrac{2}{3}x\left(x-2\right)\left(x+2\right)=0\)
mà \(\dfrac{2}{3}>0\)
nên x(x-2)(x+2)=0
hay \(\left[{}\begin{matrix}x=0\\x-2=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-2\end{matrix}\right.\)
Vậy: \(x\in\left\{0;-2;2\right\}\)
c) Ta có: \(2x\left(3x-5\right)-\left(5-3x\right)=0\)
\(\Leftrightarrow2x\left(3x-5\right)+\left(3x-5\right)=0\)
\(\Leftrightarrow\left(3x-5\right)\left(2x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-5=0\\2x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=5\\2x=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=-\dfrac{1}{2}\end{matrix}\right.\)
Vậy: \(x\in\left\{\dfrac{5}{3};-\dfrac{1}{2}\right\}\)
d) Ta có: \(\left(2x-1\right)^2-25=0\)
\(\Leftrightarrow\left(2x-1-5\right)\left(2x-1+5\right)=0\)
\(\Leftrightarrow\left(2x-6\right)\left(2x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-6=0\\2x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=6\\2x=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
Vậy: \(x\in\left\{3;-2\right\}\)
\(a)x^2-9x+20=0 \\<=>(x-4)(x-5)=0 \\<=>x=4\ hoặc\ x=5 \\b)x^2-3x-18=0 \\<=>(x+3)(x-6)=0 \\<=>x=-3\ hoặc\ x=6 \\c)2x^2-9x+9=0 \\<=>(x-3)(2x-3)=0 \\<=>x=3\ hoặc\ x=\dfrac{3}{2}\)
d: \(\Leftrightarrow3x^2-6x-2x+4=0\)
=>(x-2)(3x-2)=0
=>x=2 hoặc x=2/3
e: \(\Leftrightarrow3x\left(x^2-2x-3\right)=0\)
=>x(x-3)(x+1)=0
hay \(x\in\left\{0;3;-1\right\}\)
f: \(\Leftrightarrow x^2-5x-2+x=0\)
\(\Leftrightarrow x^2-4x-2=0\)
\(\Leftrightarrow\left(x-2\right)^2=6\)
hay \(x\in\left\{\sqrt{6}+2;-\sqrt{6}+2\right\}\)
a) \(\left(3x+5\right)\left(7-2x\right)+6x\left(x+4\right)=0\)
\(\Leftrightarrow21x-6x^2+35-10x+6x^2+24x=0\)
\(\Leftrightarrow35x=-35\Leftrightarrow x=-1\)
b) \(x^3-25x=0\)
\(\Leftrightarrow x\left(x^2-25\right)=0\)
\(\Leftrightarrow x\left(x-5\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\\x=-5\end{matrix}\right.\)
a: Ta có: \(\left(3x+5\right)\left(7-2x\right)+6x\left(x+4\right)=0\)
\(\Leftrightarrow21x-6x^2+35-10x+6x^2+24x=0\)
\(\Leftrightarrow x=1\)
b: Ta có: \(x^3-25x=0\)
\(\Leftrightarrow x\left(x-5\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\\x=-5\end{matrix}\right.\)
nhiều v~~~, dễ mà lp 8 ?