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1
20 tháng 4 2021
\(\left(x-5\right)^6=\left(x-5\right)^8\)
\(\Leftrightarrow\left(x-5\right)^6\left[1-\left(x-5\right)^2\right]=0\)
\(\Leftrightarrow\left(x-5\right)^6\left(x-4\right)\left(6-x\right)=0\Leftrightarrow x=4;x=5;x=6\)
7 tháng 5 2019
a)\(10\left(x-7\right)-8\left(x+5\right)=6\cdot\left(-5\right)+24\)
\(10x-10\cdot7-8x-8\cdot5=\left(-30\right)+24\)
\(10x-70-8x-40=-6\)
\(10x-8x=\left(-6\right)+70+40\)
\(2x=104\)
\(x=104\div2\)
\(x=52\)
b)\(2\left(4x-8\right)-7\left(3+x\right)=6\)
\(2\cdot4x-2\cdot8-7\cdot3-7x=6\)
\(8x-16-21-7x=6\)
\(8x-7x=6+16+21\)
\(x=43\)
O
18 tháng 4 2022
a) \(\left(x-\dfrac{1}{2}\right)\left(-3-\dfrac{x}{2}\right)=0\)
Th1 : \(x-\dfrac{1}{2}=0\)
\(x=0+\dfrac{1}{2}\)
\(x=\dfrac{1}{2}\)
Th2 : \(-3-\dfrac{x}{2}=0\)
\(\dfrac{x}{2}=-3\)
\(x=\left(-3\right)\cdot2\)
\(x=-6\)
Vậy \(x\) = \(\left(\dfrac{1}{2};-6\right)\)
b) \(x-\dfrac{1}{8}=\dfrac{5}{8}\)
\(x=\dfrac{5}{8}+\dfrac{1}{8}\)
\(x=\dfrac{3}{4}\)
c) \(-\dfrac{1}{2}-\left(\dfrac{3}{2}+x\right)=-2\)
\(\dfrac{3}{2}+x=-\dfrac{1}{2}-\left(-2\right)\)
\(\dfrac{3}{2}+x=\dfrac{3}{2}\)
\(x=\dfrac{3}{2}-\dfrac{3}{2}\)
\(x=0\)
d) \(x+\dfrac{1}{3}=\dfrac{-12}{5}\cdot\dfrac{10}{6}\)
\(x+\dfrac{1}{3}=-4\)
\(x=-4-\dfrac{1}{3}\)
\(x=-\dfrac{13}{3}\)
NV
Nguyễn Việt Lâm
Giáo viên
8 tháng 1
a.
\(10⋮\left(x-1\right)\)
\(\Rightarrow x-1=Ư\left(10\right)\)
\(\Rightarrow x-1=\left\{-10;-5;-2;-1;1;2;5;10\right\}\)
\(\Rightarrow x=\left\{-9;-4;-1;0;2;3;6;11\right\}\)
b.
\(\left(x+5\right)⋮\left(x-2\right)\Rightarrow\left(x-2\right)+7⋮x-2\)
\(\Rightarrow7⋮x-2\)
\(\Rightarrow x-2=Ư\left(7\right)=\left\{-7;-1;1;7\right\}\)
\(\Rightarrow x=\left\{-5;1;3;9\right\}\)
c.
\(\left(3x+8\right)⋮\left(x-1\right)\)
\(\Rightarrow\left(3x-3+11\right)⋮\left(x-1\right)\)
\(\Rightarrow3\left(x-1\right)+11⋮x-1\)
\(\Rightarrow11⋮\left(x-1\right)\)
\(\Rightarrow x-1=Ư\left(11\right)=\left\{-11;-1;1;11\right\}\)
\(\Rightarrow x=\left\{-10;0;2;12\right\}\)
20 tháng 9 2023
a) \(\left(x-1\right)^3=8=2^3\)
\(x-1=2\)
\(x=2+1=3\)
b) \(7^{2x-6}=49=7^2\)
\(2x-6=2\)
\(2x=6+2=8\)
\(x=8:2=4\)
c) \(\left(2x-14\right)^7=128=2^7\)
\(2x-14=2\)
\(2x=14+2=16\)
\(x=16:2=8\)
d) \(x^4\cdot x^5=5^3\cdot5^6=5^4\cdot5^5\)
\(x=5\)
e) \(3\cdot\left(x+2\right):7\cdot4=120\)
\(x+2=120:3\cdot7:4\)
\(x+2=70\)
\(x=70-2=68\)
AH
Akai Haruma
Giáo viên
20 tháng 9 2023
Lời giải:
a. $(x-1)^3=8=2^3$
$\Rightarrow x-1=2$
$\Rightarrow x=3$
b. $7^{2x-6}=49=7^2$
$\Rightarrow 2x-6=2$
$\Rightarrow 2x=8$
$\Rightarrow x=4$
c. $(2x-14)^7=128=2^7$
$\Rightarrow 2x-14=2$
$\Rightarrow 2x=16$
$\Rightarrow x=18$
d.
$x^4.x^5=5^3.5^6$
$x^9=5^9$
$\Rightarrow x=5$
e.
$3(x+2):7=120:4=30$
$3(x+2)=30.7=210$
$x+2=210:3=70$
$x=70-2=68$
1 tháng 9 2021
giúp mik mik đang cần gấp
nhưng phả có lời giải đừng cho mỗi đáp án
1 tháng 9 2021
a:Ta có: \(\left(x-9\right)^7=\left(x-9\right)^4\)
\(\Leftrightarrow\left(x-9\right)^4\cdot\left[\left(x-9\right)^3-1\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-9=0\\x-9=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=9\\x=10\end{matrix}\right.\)
b: ta có: \(\left(3x-15\right)^{15}=\left(3x-15\right)^{10}\)
\(\Leftrightarrow\left(3x-15\right)^{10}\cdot\left[\left(3x-15\right)^5-1\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-15=0\\3x-15=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{16}{3}\end{matrix}\right.\)
22 tháng 2 2022
a: =>2x-1=-2
=>2x=-1
hay x=-1/2
b: \(\Leftrightarrow\left[{}\begin{matrix}3x+2=0\\-\dfrac{2}{5}x-7=0\end{matrix}\right.\Leftrightarrow x\in\left\{-\dfrac{2}{3};-\dfrac{35}{2}\right\}\)
c: x/8=9/4
nên x/8=18/8
hay x=18
d: \(\Leftrightarrow\left(x-3\right)^2=36\)
=>x-3=6 hoặc x-3=-6
=>x=9 hoặc x=-3
e: =>-1,7x=6,12
hay x=-3,6
h: =>x-3,4=27,6
hay x=31
R
22 tháng 2 2022
a) \(\dfrac{1}{3}\div\left(2x-1\right)=\dfrac{-1}{6}\)
\(\left(2x-1\right).\dfrac{1}{3}\div\left(2x-1\right)=\left(2x-1\right)\left(-\dfrac{1}{6}\right)\)
\(\dfrac{1}{3}=\left(2x-1\right)\left(-\dfrac{1}{6}\right)\)
\(\dfrac{1}{3}=-1\left(2x-1\right)\div6\)
\(\dfrac{1}{3}=-2x+1\div6\)
\(x=-\dfrac{1}{2}\)
b) \(\left(3x+2\right)\left(\dfrac{-2}{5}x-7\right)=0\)
\(TH1:3x+2=0\)
\(3x=0-2\)
\(3x=-2\)
\(x=\dfrac{-2}{3}\)
\(TH2:\left(-\dfrac{2}{5}x-7\right)=0\)
\(\left(\dfrac{-2}{5}x-7\right)=0\)
\(\left(\dfrac{-2x}{5}+\dfrac{5\left(-7\right)}{5}\right)=0\)
\(\left(\dfrac{-2x-35}{5}\right)=0\)
\(-2x-35=0\)
\(-2x=0+35\)
\(x=-\dfrac{35}{2}\)
c) \(\dfrac{x}{8}=\dfrac{9}{4}\)
\(\Leftrightarrow x=\dfrac{9.8}{4}=\dfrac{72}{4}=18\)
\(x=18\)
d) \(\dfrac{x-3}{2}=\dfrac{18}{x-3}\)
\(x-3=18+2\)
\(x=20-3\)
\(x=17\)
e) \(4,5x-6,2x=6,12\)
\(\dfrac{9x}{2}-6,2.x=6,12\)
\(\dfrac{9x}{2}+\dfrac{-31x}{5}=6,12\)
\(\dfrac{5.9x}{10}+\dfrac{2\left(-31\right)x}{10}=6.12\)
\(\dfrac{45x-62x}{10}=6.12\)
\(=-17x\div10=6.12\)
\(-17x=10.6.12\)
\(x=-3,6\)
h) \(11,4-\left(x-3,4\right)=-16,2\)
\(x-3,4=-16,2+11,4\)
\(x-3,4=-4,8\)
\(x=-1,4\)