tìm X
I x - 3 I = 2x + 4
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TP
0
13 tháng 6 2021
`4|8x-3|+5|3-2x|=18`
`<=>4|8x-3|+5|2x-3|=18`
Nếu `x>=3/2=>|8x-3|=8x-3,|2x-3|=2x-3`
`pt<=>4(8x-3)+5(2x-3)=18`
`<=>32x-12+10x-15=18`
`<=>42x=27+18=45`
`<=>x=45/42(l)`
Nếu `x<=3/8=>|8x-3|=3-8x,|2x-3|=3-2x`
`pt<=>4(3-8x)+5(3-2x)=18`
`<=>12-32x+15-10x=18`
`<=>27-42x=18`
`<=>9=42x`
`<=>x=3/14(tm)`
Nếu `3/9<=x<=3/2=>|8x-3|=8x-3,|2x-3|=3-2x`
`pt<=>4(8x-3)+5(3-2x)=18`
`<=>32x-12+15-10x=18`
`<=>22x+3=18`
`<=>22x=15`
`<=>x=15/22(tm)`
Vậy x=15/22 hoặc x=3/14
19 tháng 1 2017
b) Theo bài ra , ta có :
(2x - 5) - (3x - 7) = x + 3
(=) 2x - 5 - 3x + 7 = x + 3
(=) -2x = 1
(=) x = -1/2
Vậy x = -1/2
Chúc bạn học tốt =))
28 tháng 10 2023
a: \(\left(\dfrac{1}{4}-x\right)\left(x+\dfrac{2}{5}\right)=0\)
=>\(\left[{}\begin{matrix}\dfrac{1}{4}-x=0\\x+\dfrac{2}{5}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{4}\\x=-\dfrac{2}{5}\end{matrix}\right.\)
b: \(\left|2x+1\right|+\dfrac{3}{2}=2\)
=>\(\left|2x+1\right|=\dfrac{1}{2}\)
=>\(\left[{}\begin{matrix}2x+1=\dfrac{1}{2}\\2x+1=-\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=-\dfrac{1}{2}\\2x=-\dfrac{3}{2}\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=-\dfrac{1}{4}\\x=-\dfrac{3}{4}\end{matrix}\right.\)
c: (2x-3)2=36
=>\(\left[{}\begin{matrix}2x-3=6\\2x-3=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=9\\2x=-3\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=\dfrac{9}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)
d: \(7^{x+2}+2\cdot7^x=357\)
=>\(7^x\cdot49+7^x\cdot2=357\)
=>\(7^x=7\)
=>x=1
T
28 tháng 10 2023
a) \(\left(\dfrac{1}{4}-x\right)\left(x+\dfrac{2}{5}\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}\dfrac{1}{4}-x=0\\x+\dfrac{2}{5}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{4}\\x=-\dfrac{2}{5}\end{matrix}\right.\)
\(---\)
b) \(\left|2x+1\right| +\dfrac{2}{3}=2\)
\( \Rightarrow\left|2x+1\right|=2-\dfrac{2}{3}\)
\(\Rightarrow\left|2x+1\right|=\dfrac{4}{3}\)
\(\Rightarrow\left[{}\begin{matrix}2x+1=\dfrac{4}{3}\\2x+1=-\dfrac{4}{3}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=\dfrac{1}{3}\\2x=-\dfrac{7}{3}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{6}\\x=-\dfrac{7}{6}\end{matrix}\right.\)
\(---\)
c) \(\left(2x-3\right)^2=36\)
\(\Rightarrow\left(2x-3\right)^2=\left(\pm6\right)^2\)
\(\Rightarrow\left[{}\begin{matrix}2x-3=6\\2x-3=-6\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=9\\2x=-3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{9}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)
\(---\)
d) \(7^{x+2}+2\cdot7^x=357\)
\(\Rightarrow7^x\cdot7^2+2\cdot7^x=357\)
\(\Rightarrow7^x\cdot\left(7^2+2\right)=357\)
\(\Rightarrow7^x\cdot\left(49+2\right)=357\)
\(\Rightarrow7^x\cdot51=357\)
\(\Rightarrow7^x=357:51\)
\(\Rightarrow7^x=7\)
\(\Rightarrow x=1\)
Ta có hai trường hợp:
TH1: x-3=2x+4
-x=7
x=-7
TH2 : x-3=-2x-4
3x=-1
x=-1/3
\(\left|x-3\right|=2x+4\)
TH1: \(x-3=2x+4\)
\(\Rightarrow x-2x=4+3\)
\(x.\left(1-2\right)=7\)
\(-1x=7\)
\(x=-7\)
TH2: \(x-3=-\left(2x+4\right)\)
\(x-3=-2x-4\)
\(x+2x=-4+3\)
\(x.\left(1+2\right)=-1\)
\(x3=-1\)
\(x=\frac{-1}{3}\)
KL: x= -7; x= -1/3
CHÚC BN HỌC TỐT!!!