Tìm x
.|x−3|−2|5−2x|=11
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\(a,3\left(2x-3\right)+2\left(2-x\right)=-3\\ \Leftrightarrow6x-9+4-2x=-3\\ \Leftrightarrow4x=2\\ \Leftrightarrow x=\dfrac{1}{2}\\ b,x\left(5-2x\right)+2x\left(x-1\right)=13\\ \Leftrightarrow5x-2x^2+2x^2-2x=13\\ \Leftrightarrow3x=13\\ \Leftrightarrow x=\dfrac{13}{3}\\ c,5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\\ \Leftrightarrow5x^2-5x-5x^2-3x+14=6\\ \Leftrightarrow-8x=-8\\ \Leftrightarrow x=1\\ d,3x\left(2x+3\right)-\left(2x+5\right)\left(3x-2\right)=8\\ \Leftrightarrow6x^2+9x-6x^2-11x+10=8\\ \Leftrightarrow-2x=-2\\ \Leftrightarrow x=1\)
\(e,2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\\ \Leftrightarrow10x-16-12x+15=12x-16+11\\ \Leftrightarrow-14x=-4\\ \Leftrightarrow x=\dfrac{2}{7}\\ f,2x\left(6x-2x^2\right)+3x^2\left(x-4\right)=8\\ \Leftrightarrow12x^2-4x^3+3x^3-12x^2=8\\ \Leftrightarrow-x^3-8=0\\ \Leftrightarrow-\left(x^3+8\right)=0\\ \Leftrightarrow-\left(x+2\right)\left(x^2-2x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-2\\x\in\varnothing\left(x^2-2x+4=\left(x-1\right)^2+3>0\right)\end{matrix}\right.\)
Bài 4:
a: Ta có: \(3\left(2x-3\right)-2\left(x-2\right)=-3\)
\(\Leftrightarrow6x-9-2x+4=-3\)
\(\Leftrightarrow4x=2\)
hay \(x=\dfrac{1}{2}\)
b: Ta có: \(x\left(5-2x\right)+2x\left(x-1\right)=13\)
\(\Leftrightarrow5x-2x^2+2x^2-2x=13\)
\(\Leftrightarrow3x=13\)
hay \(x=\dfrac{13}{3}\)
c: Ta có: \(5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\)
\(\Leftrightarrow5x^2-5x-5x^2+7x-10x+14=6\)
\(\Leftrightarrow-8x=-8\)
hay x=1
`2//(5x-8)-3(4x-5)=4(3x-4)`
`<=>5x-8-12x+15=12x-16`
`<=>-19x=-23`
`<=>x=23/19` Vậy `x=23/19`
`3//2(x^3-1)-2x^2(x+2x^4)+(4x^5+4)x=6`
`<=>2x^3-2-2x^3-4x^6+4x^6+4x=6`
`<=>4x=8`
`<=>x=2` Vậy `x=2`
3(x+5) -3 = 2(x+1) +7
⇒3x+15-3=2x+2+7
⇒3x+12=2x+9
⇒x=-3
15-(x+2)² = -1
⇒(x+2)2=16
⇒\(\left[{}\begin{matrix}x+2=4\\x+2=-4\end{matrix}\right.\)
⇒\(\left[{}\begin{matrix}x=2\\x=-6\end{matrix}\right.\)
75-(2x+1)³ =11
⇒(2x+1)³=64⇒\(\left[{}\begin{matrix}2x+1=8\\2x+1=-8\end{matrix}\right.\)⇒\(\left[{}\begin{matrix}x=3,5\\x=-4,5\end{matrix}\right.\)a) ta có: 3(x+5)-3=2(x+1)+7
nên 3x-2x=2+7-15+3
hay x=-3
b) Ta có: \(15-\left(x+2\right)^2=-1\)
nên \(\left(x+2\right)^2=16\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=4\\x+2=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-6\end{matrix}\right.\)
Tìm x biết:
a,x-5/7=1/9
b,2x/5=6/2x+1
c,11/8+13/6=85/x
d,2x-2/11=1.1/5
e,x/15=3/5+-2/3
f,x/182=-6/14.35/91
a, \(x\) - \(\dfrac{5}{7}\) = \(\dfrac{1}{9}\)
\(x\) = \(\dfrac{1}{9}\) + \(\dfrac{5}{7}\)
\(x\) = \(\dfrac{52}{63}\)
b, \(\dfrac{2x}{5}\) = \(\dfrac{6}{2x+1}\)
2\(x\).(2\(x\) + 1) = 30
4\(x^2\)+ 2\(x\) - 30 = 0
4\(x^2\) + 12\(x\) - 10\(x\) - 30 = 0
(4\(x^2\) + 12\(x\)) - (10\(x\) + 30) =0
4\(x\).(\(x\) + 3) - 10.(\(x\) +3) = 0
2 (\(x\) + 3).(2\(x\) - 5) = 0
\(\left[{}\begin{matrix}x+3=0\\2x-5=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=-3\\x=\dfrac{5}{2}\end{matrix}\right.\)
Vậy \(x\) \(\in\) {-3; \(\dfrac{5}{2}\)}
a) Ta có: \(3\left(x+5\right)-3=2\left(x+1\right)+7\)
\(\Leftrightarrow3x+15-3=2x+2+7\)
\(\Leftrightarrow3x+12=2x+9\)
hay x=-3
b) Ta có: \(15-\left(x+2\right)^2=-1\)
\(\Leftrightarrow\left(x+2\right)^2=16\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=4\\x+2=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-6\end{matrix}\right.\)
c) Ta có: \(75-\left(2x+1\right)^3=11\)
\(\Leftrightarrow\left(2x+1\right)^3=64\)
\(\Leftrightarrow2x+1=4\)
hay \(x=\dfrac{3}{2}\)
`3(x+5)-3=2(x+1)+7`
`3x+15-3=2x+2+7`
`x=-3`
.
`15-(x+2)^2=-1`
`(x+2)^2=16`
`(x+2)^2=4^2=(-4)^2`
`[(x+2=4),(x+2=-4):}`
`[(x=2),(x=-6):}`
.
`75-(2x+1)^3=11`
`(2x+1)^3=64`
`(2x+1)^2=4^3`
`2x+1=4`
`x=3/2`
5^x + 5^ ( x + 2 ) = 650
5x + 5x . 52 = 650
5x .( 1 + 25 ) = 650
5x . 26 = 650
5x = 650 : 26
5x = 25
5x = 52
=> x = 2
Vậy x = 2
Lời giải:
Nếu $x\geq 3$ thì:
$x-3-2(2x-5)=11$
$-3x+7=11$
$-3x=4$
$x=\frac{-4}{3}< 3$ (loại)
Nếu $3> x\geq \frac{5}{2}$ thì:
$3-x-2(2x-5)=11$
$13-5x=11$
$5x=2$
$x=\frac{2}{5}$< \frac{5}{2}$ (loại)
Nếu $x< \frac{5}{2}$ thì:
$3-x-2(5-2x)=11$
$3x-7=11$
$3x=18$
$x=6> \frac{5}{2}$
Vậy không tồn tại $x$ thỏa mãn