tìm x biết:
A= 5*x-3*(4*x-2*(4*x-3*(5*x-2)))=182
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`@` ` \text {Ans}`
`\downarrow`
`a,`
`1/4+3/4*x=3/2-x`
`=> 1/4 + 3/4x - 3/2 + x = 0`
`=> (1/4 - 3/2) + (3/4x + x) = 0`
`=> -5/4 + 7/4x = 0`
`=> 7/4x = 5/4`
`=> x = 5/4 \div 7/4`
`=> x = 5/7`
Vậy, `x=5/7`
`b,`
`3/5*x-1/4=1/10*x-1/2`
`=> 3/5x - 1/4 - 1/10x + 1/2 = 0`
`=> (3/5x - 1/10x) + (-1/4 + 1/2)=0`
`=> 1/2x + 1/4 = 0`
`=> 1/2x = -1/4`
`=> x = -1/4 \div 1/2`
`=> x = -1/2`
Vậy, `x=-1/2`
`c,`
`3x-3/5=x-1/4`
`=> 3x - 3/5 - x + 1/4 = 0`
`=> (3x - x) - (3/5 - 1/4) = 0`
`=> 2x - 7/20 = 0`
`=> 2x = 0,35`
`=> x = 0,35 \div 2`
`=> x = 7/40`
Vậy, `x=7/40`
`d,`
`3/2*x-2/5=1/3*x-1/4`
`=> 3/2x - 2/5 - 1/3x + 1/4 = 0`
`=> (3/2x - 1/3x) - (2/5 - 1/4) = 0`
`=> 7/6x - 3/20 = 0`
`=> 7/6x = 3/20`
`=> x = 3/20 \div 7/6`
`=> x = 9/70`
Vậy, `x=9/70`
`@` `\text {Kaizuu lv uuu}`
a) \(\left(x+3\right)\left(2x-1\right)-\left(x-3\right)\left(x+1\right)=0\)
\(\Leftrightarrow2x^2+5x-3-x^2+2x+3=0\)
\(\Leftrightarrow x^2+7x=0\Leftrightarrow x\left(x+7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-7\end{matrix}\right.\)
b) \(\left(x+4\right)\left(2x-3\right)-3\left(x-2\right)\left(x+2\right)=0\)
\(\Leftrightarrow2x^2+5x-12-3x^2+12=0\)
\(\Leftrightarrow x^2-5x=0\Leftrightarrow x\left(x-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\end{matrix}\right.\)
a) \(\Rightarrow3x\left(x-5\right)-2\left(x-5\right)=0\)
\(\Rightarrow\left(x-5\right)\left(3x-2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{2}{3}\end{matrix}\right.\)
b) \(\Rightarrow x^3+6x^2+12x+8-x^3+6x^2=4\)
\(\Rightarrow12x^2+12x+4=0\)
\(\Rightarrow x\in\varnothing\)(do \(12x^2+12x+4=12\left(x^2+x+\dfrac{1}{4}\right)+1=12\left(x+\dfrac{1}{2}\right)^2+1\ge1>0\))
a: =>3x+3=5x-25
=>-2x=-28
hay x=14
b: =>3x+6=-4x+20
=>7x=14
hay x=2
2(x4+3)-(9)=17
⇒2x4+6+9=17
⇒2x4+15=17
⇒ 2x4=2
⇒ x4=1
⇒ x=\(\pm1\)
b) 5x2.x+1-3.42=-47
⇒5x3+1-48=-47
⇒5x3-47=-47
⇒5x3=0
⇒x3=0
⇒x=0
a) \(2\left(x^4+3\right)-\left(-9\right)=17\)
\(2x^4+6+9=17\)
\(2x^4=2\)
\(x^4=1\)
⇒ \(x=1\)
\((2x-1)^2+(x+3)^2-5(x+7)(x-7)=0\)
\(< =>4x^2-4x+1+x^2+6x+9-5\left(x^2-7^2\right)=0\\ < =>4x^2-4x+1+x^2+6x+9-5x^2+245=0\\ < =>2x+255=0\\ < =>2x=-255=>x=\dfrac{-255}{2}\)
Vậy \(x=\dfrac{-255}{2}\)
\(\Rightarrow4x^2-4x+1+x^2+6x+9-5x^2+245=0\)
\(\Rightarrow2x+255=0\Rightarrow2x=-255\Rightarrow x=-\dfrac{255}{2}\)