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a) \(\left(x-\frac{2}{7}\right)^6=0\)

\(\Rightarrow x-\frac{2}{7}=0\)

\(\Rightarrow x=\frac{2}{7}\)

b) \(\frac{2^{4-x}}{16}=32^6\)

\(\Rightarrow\frac{2^{4-x}}{2^4}=\left(2^5\right)^6\)

\(\Rightarrow2^{-x}=2^{30}\)

=> -x = 30

=> x = -30

30 tháng 6 2021

Em hc bảng xét dáu chx ??

30 tháng 6 2021

Lớp 7 chưa học đâu em

28 tháng 8 2018

\(a,\frac{x+5}{2010}+\frac{x+6}{2009}+\frac{x+7}{2008}=-3\)

\(\Rightarrow\left(\frac{x+5}{2010}+1\right)+\left(\frac{x+6}{2009}+1\right)+\left(\frac{x+7}{2008}+1\right)=0\)

\(\Rightarrow\frac{x+2016}{2010}+\frac{x+2016}{2009}+\frac{x+2006}{2008}=0\)

chỉ bt lm v thoi "(

28 tháng 8 2018

a)   \(\frac{x+5}{2010}+\frac{x+6}{2009}+\frac{x+7}{2008}=-3\)

<=>   \(\frac{x+5}{2010}+1+\frac{x+6}{2009}+1+\frac{x+7}{2008}+1=0\)

<=>  \(\frac{x+2015}{2010}+\frac{x+2015}{2009}+\frac{x+2015}{2008}=0\)

<=>  \(\left(x+2015\right)\left(\frac{1}{2010}+\frac{1}{2009}+\frac{1}{2008}\right)=0\)

<=> \(x+2015=0\)    (do  1/2010 + 1/2009 + 1/2008 # 0 )

<=>   \(x=-2015\)

Vậy...

b)  mạo phép chỉnh đề

   \(\frac{x+2}{327}+\frac{x+3}{326}+\frac{x+4}{325}+\frac{x+344}{5}=0\)

<=>  \(\frac{x+2}{327}+1+\frac{x+3}{326}+1+\frac{x+4}{325}+1+\frac{x+344}{5}-3=0\)

<=>  \(\frac{x+329}{327}+\frac{x+329}{326}+\frac{x+329}{325}+\frac{x+329}{5}=0\)

làm tương tự a

16 tháng 10 2021

\(1,\\ a,\Leftrightarrow x-\dfrac{1}{3}=0\Leftrightarrow x=\dfrac{1}{3}\\ b,\Leftrightarrow\left[{}\begin{matrix}x-4=4\\x-4=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\x=0\end{matrix}\right.\\ c,\Leftrightarrow2x+1=-2\Leftrightarrow x=-\dfrac{3}{2}\\ 2,\\ a,=1\\ b,=\left(\dfrac{13}{4}\right)^2=\dfrac{169}{16}\\ c,=\left(-\dfrac{7}{4}\right)^2=\dfrac{49}{16}\\ d,=\left(\dfrac{3}{7}\right)^{20}:\left(\dfrac{3}{7}\right)^{12}=\left(\dfrac{3}{7}\right)^8=...\\ e,=\left(3\cdot5\cdot\dfrac{2}{3}\right)^2=10^2=100\)

23 tháng 9 2021

a) \(\Leftrightarrow x-\dfrac{1}{2}=0\Leftrightarrow x=\dfrac{1}{2}\)

b) \(\Leftrightarrow\left[{}\begin{matrix}x-2=1\\x-2=-1\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-1\end{matrix}\right.\)

c) \(\Leftrightarrow\left(2x-1\right)^3=\left(-2\right)^3\)

\(\Leftrightarrow2x-1=-3\Leftrightarrow x=-1\)

d) \(\left(x+\dfrac{1}{2}\right)^2=\dfrac{1}{6}\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{\sqrt{6}}{6}\\x+\dfrac{1}{2}=-\dfrac{\sqrt{6}}{6}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-3+\sqrt{6}}{6}\\x=-\dfrac{3+\sqrt{6}}{6}\end{matrix}\right.\)

b: Ta có: \(\left(x-2\right)^2=1\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=1\\x-2=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)

c: Ta có: \(\left(2x-1\right)^3=-8\)

\(\Leftrightarrow2x-1=-2\)

\(\Leftrightarrow2x=-1\)

hay \(x=-\dfrac{1}{2}\)

7 tháng 12 2021

\(1,\dfrac{x}{2}=\dfrac{y}{5}=\dfrac{x+y}{2+5}=\dfrac{21}{7}=3\\ \Rightarrow\left\{{}\begin{matrix}x=6\\y=15\end{matrix}\right.\\ 2,7x=3y\Rightarrow\dfrac{x}{3}=\dfrac{y}{7}=\dfrac{x-y}{3-7}=\dfrac{16}{-4}=-4\\ \Rightarrow\left\{{}\begin{matrix}x=-12\\y=-28\end{matrix}\right.\\ 3,\dfrac{x}{5}=\dfrac{y}{6}=\dfrac{z}{7}=\dfrac{x-y-z}{5-6-7}=\dfrac{36}{-8}=-\dfrac{9}{2}\\ \Rightarrow\left\{{}\begin{matrix}x=-\dfrac{45}{2}\\y=-27\\z=-\dfrac{63}{2}\end{matrix}\right.\\ 4,x:y:z=3:5:7\Rightarrow\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{7}=\dfrac{2x+3y-z}{6+15-7}=\dfrac{-14}{14}=-1\\ \Rightarrow\left\{{}\begin{matrix}x=-3\\y=-5\\z=-7\end{matrix}\right.\)

7 tháng 12 2021

3. Áp dụng tính chất dãy tỉ số bằng nhau, ta có:

\(\dfrac{x}{5}=\dfrac{y}{6}=\dfrac{z}{7}=\dfrac{x-y-z}{5-6-7}=\dfrac{36}{-8}=\dfrac{-9}{2}\)

\(x=\dfrac{-45}{2}\)

\(y=-27\)

\(z=\dfrac{-63}{2}\)

11 tháng 10 2021

Làm vs mn cần gấp

 

11 tháng 10 2021

\(a,\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{1}{5}=0\\\dfrac{8}{5}+2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=\dfrac{4}{5}\end{matrix}\right.\)

\(b,\dfrac{x-\dfrac{4}{7}}{x+\dfrac{1}{2}}>0\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-\dfrac{4}{7}>0\\x+\dfrac{1}{2}>0\end{matrix}\right.\\\left\{{}\begin{matrix}x-\dfrac{4}{7}< 0\\x+\dfrac{1}{2}< 0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x>\dfrac{4}{7}\\x< -\dfrac{1}{2}\end{matrix}\right.\)

\(c,\dfrac{2x-3}{x+\dfrac{7}{4}}< 0\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}2x-3< 0\\x+\dfrac{7}{4}>0\end{matrix}\right.\\\left\{{}\begin{matrix}2x-3>0\\x+\dfrac{7}{4}< 0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x< \dfrac{3}{2}\\x >-\dfrac{7}{4}\end{matrix}\right.\\\left\{{}\begin{matrix}x>\dfrac{3}{2}\\x< -\dfrac{7}{4}\end{matrix}\right.\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}-\dfrac{7}{4}< x< \dfrac{3}{2}\\x\in\varnothing\end{matrix}\right.\Leftrightarrow-\dfrac{7}{4}< x< \dfrac{3}{2}\)

Bài 2: 

a: =>x=0 hoặc x=-3

b: =>x-2=0 hoặc 5-x=0

=>x=2 hoặc x=5

c: =>x-1=0

hay x=1

8 tháng 9 2021

\(a,\left|x+\dfrac{4}{5}\right|-\dfrac{1}{7}=0\\ \Rightarrow\left|x+\dfrac{4}{5}\right|=\dfrac{1}{7}\\ \Rightarrow\left[{}\begin{matrix}x+\dfrac{4}{5}=\dfrac{1}{7},\forall x+\dfrac{4}{5}\ge0\\x+\dfrac{4}{5}=-\dfrac{1}{7},\forall x+\dfrac{4}{5}< 0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{23}{35},\forall x\ge-\dfrac{4}{5}\left(N\right)\\x=-\dfrac{33}{35},\forall x< -\dfrac{4}{5}\left(N\right)\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{23}{35}\\x=-\dfrac{33}{35}\end{matrix}\right.\)

\(b,\left|x-2\right|=x-2\\ \Rightarrow\left[{}\begin{matrix}x-2=x-2,\forall x-2\ge0\\x-2=2-x,\forall x-2< 0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=0,\forall x\ge2\left(L\right)\\x=2,\forall x< 2\left(L\right)\end{matrix}\right.\\ \Rightarrow x\in\varnothing\)

d: Ta có: (-4,6)+x=-3,5

nên x=-3,5+4,6

hay x=1,1

a: Ta có: \(1.5-\left|x-0.3\right|=0\)

\(\Leftrightarrow\left|x-0.3\right|=1.5\)

\(\Leftrightarrow\left[{}\begin{matrix}x-0.3=1.5\\x-0.3=-1.5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1.8\\x=-1.2\end{matrix}\right.\)

a: =>7(x-5)>0

=>x-5>0

=>x>5

b: =>x-1 thuộc {1;-1;11;-11}

=>x thuộc {2;0;12;-10}

c: =>x+1+7 chia hết cho x+1

=>x+1 thuộc {1;-1;7;-7}

=>x thuộc {0;-2;6;-8}

d: =>(x+2)(x-5)<0

=>-2<x<5

30 tháng 7 2023

a:(- 7) . ( 5 – x) < 0

=>7(x-5)>0

=>x-5>0

=>x>5

b:11 ⁝ x – 1

=>x-1 thuộc {1;-1;11;-11}

=>x thuộc {2;0;12;-10}

c: x + 8 ⁝ x + 1

=>x+1+7 chia hết cho x+1

=>x+1 thuộc {1;-1;7;-7}

=>x thuộc {0;-2;6;-8}

d: (x + 2) . (5 – x) > 0

=>(x+2)(x-5)<0

=>-2<x<5