K
Khách

Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

21 tháng 9 2021

\(c,\Rightarrow\left[{}\begin{matrix}-2\left(x+2\right)+\left(4-x\right)=11\left(x< -2\right)\\2\left(x+2\right)+\left(4-x\right)=11\left(-2\le x\le4\right)\\2\left(x+2\right)+\left(x-4\right)=11\left(x>4\right)\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{11}{3}\left(tm\right)\\x=3\left(tm\right)\\x=\dfrac{11}{3}\left(ktm\right)\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{11}{3}\end{matrix}\right.\)

21 tháng 9 2021

\(a,\Rightarrow\left[{}\begin{matrix}x+\dfrac{5}{2}=3x+1\\x+\dfrac{5}{2}=-3x-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=-\dfrac{7}{8}\end{matrix}\right.\)

 

23 tháng 11 2021

\(a,\Rightarrow\left[{}\begin{matrix}5x+1=\dfrac{6}{7}\\5x+1=-\dfrac{6}{7}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}5x=\dfrac{1}{7}\\5x=-\dfrac{13}{7}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{35}\\x=-\dfrac{13}{35}\end{matrix}\right.\\ b,\Rightarrow\left(-\dfrac{1}{8}\right)^x=\dfrac{1}{64}=\left(-\dfrac{1}{8}\right)^2\Rightarrow x=2\\ c,\Rightarrow\left(x-2\right)\left(2x+3\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=2\\x=-\dfrac{3}{2}\end{matrix}\right.\\ d,\Rightarrow\left(x+1\right)^{x+10}-\left(x+1\right)^{x+4}=0\\ \Rightarrow\left(x+1\right)^{x+4}\left[\left(x+1\right)^6-1\right]=0\\ \Rightarrow\left[{}\begin{matrix}x+1=0\\\left(x+1\right)^6=1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x+1=0\\x+1=1\\x+1=-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-1\\x=0\\x=-2\end{matrix}\right.\\ e,\Rightarrow\dfrac{3}{4}\sqrt{x}=\dfrac{5}{6}\left(x\ge0\right)\\ \Rightarrow\sqrt{x}=\dfrac{10}{9}\Rightarrow x=\dfrac{100}{81}\)

a) (x-1):2/3=-2/5

=>x-1=-4/15

=>x=11/15

b) |x-1/2|-1/3=0

=>|x-1/2|=1/3

=>\(\left\{{}\begin{matrix}x=\dfrac{1}{3}+\dfrac{1}{2}=\dfrac{5}{6}\\x=-\dfrac{1}{3}+\dfrac{1}{2}=\dfrac{1}{6}\end{matrix}\right.\) 

c) Tương Tự câu B

 

6 tháng 12 2023

\(a,1-3\left|2x-3\right|=-\dfrac{1}{2}\\ 3\left|2x-3\right|=1+\dfrac{1}{2}\\ 3\left|2x-3\right|=\dfrac{3}{2}\\ \left|2x-3\right|=\dfrac{3}{2}:3\\ \left|2x-3\right|=\dfrac{9}{2}\\ \Rightarrow\left[{}\begin{matrix}2x-3=\dfrac{9}{2}\\2x-3=-\dfrac{9}{2}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=\dfrac{15}{2}\\2x=-\dfrac{3}{2}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{15}{4}\\x=-\dfrac{3}{4}\end{matrix}\right.\)

Vậy `x in {15/4;-3/4}`

\(b,\left(\left|x\right|-0,2\right)\left(x^3-8\right)=0\\ \left(\left|x\right|-0,2\right)\left(x-2\right)\left(x^2+2x+4\right)=0\\ \Rightarrow\left[{}\begin{matrix}\left|x\right|-0,2=0\\x-2=0\\x^2+2x+4=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}\left|x\right|=0,2\\x=2\\\left(x+1\right)^2+3=0\left(lọai\right)\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=0,2\\x=-0,2\\x=2\end{matrix}\right.\)

Vậy `x in {+-0,2;2}`

1 tháng 9 2023

\(a,3-x=x+1,8\)

\(\Rightarrow-x-x=1,8-3\)

\(\Rightarrow-2x=-1,2\)

\(\Rightarrow x=0,6\)

\(b,2x-5=7x+35\)

\(\Rightarrow2x-7x=35+5\)

\(\Rightarrow-5x=40\)

\(\Rightarrow x=-8\)

\(c,2\left(x+10\right)=3\left(x-6\right)\)

\(\Rightarrow2x+20=3x-18\)

\(\Rightarrow2x-3x=-18-20\)

\(\Rightarrow-x=-38\)

\(\Rightarrow x=38\)

\(d,8\left(x-\dfrac{3}{8}\right)+1=6\left(\dfrac{1}{6}+x\right)+x\)

\(\Rightarrow8x-3+1=1+6x+x\)

\(\Rightarrow8x-3=7x\)

\(\Rightarrow8x-7x=3\)

\(\Rightarrow x=3\)

\(e,\dfrac{2}{9}-3x=\dfrac{4}{3}-x\)

\(\Rightarrow-3x+x=\dfrac{4}{3}-\dfrac{2}{9}\)

\(\Rightarrow-2x=\dfrac{10}{9}\)

\(\Rightarrow x=-\dfrac{5}{9}\)

1 tháng 9 2023

\(g,\dfrac{1}{2}x+\dfrac{5}{6}=\dfrac{3}{4}x-\dfrac{1}{2}\)

\(\Rightarrow\dfrac{1}{2}x-\dfrac{3}{4}x=-\dfrac{1}{2}-\dfrac{5}{6}\)

\(\Rightarrow-\dfrac{1}{4}x=-\dfrac{4}{3}\)

\(\Rightarrow x=\dfrac{16}{3}\)

\(h,x-4=\dfrac{5}{6}\left(6-\dfrac{6}{5}x\right)\)

\(\Rightarrow x-4=5-x\)

\(\Rightarrow x+x=5+4\)

\(\Rightarrow2x=9\)

\(\Rightarrow x=\dfrac{9}{2}\)

\(k,7x^2-11=6x^2-2\)

\(\Rightarrow7x^2-6x^2=-2+11\)

\(\Rightarrow x^2=9\Rightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)

\(m,5\left(x+3\cdot2^3\right)=10^2\)

\(\Rightarrow5\left(x+24\right)=100\)

\(\Rightarrow x+24=20\)

\(\Rightarrow x=-4\)

\(n,\dfrac{4}{9}-\left(\dfrac{1}{6^2}\right)=\dfrac{2}{3}\left(x-\dfrac{2}{3}\right)^2+\dfrac{5}{12}\)

\(\Rightarrow\dfrac{2}{3}\left(x-\dfrac{2}{3}\right)^2+\dfrac{5}{12}=\dfrac{4}{9}-\dfrac{1}{36}\)

\(\Rightarrow\dfrac{2}{3}\left(x-\dfrac{2}{3}\right)^2+\dfrac{5}{12}=\dfrac{5}{12}\)

\(\Rightarrow\dfrac{2}{3}\left(x-\dfrac{2}{3}\right)^2=0\)

\(\Rightarrow x-\dfrac{2}{3}=0\Rightarrow x=\dfrac{2}{3}\)

#\(Urushi\text{☕}\)

a: \(-\dfrac{3}{2}x+\dfrac{1}{4}=\dfrac{1}{2}\left(x+1\right)\)

=>\(-\dfrac{3}{2}x+\dfrac{1}{4}=\dfrac{1}{2}x+\dfrac{1}{2}\)

=>\(-\dfrac{3}{2}x-\dfrac{1}{2}x=\dfrac{1}{2}-\dfrac{1}{4}\)

=>\(-2x=\dfrac{1}{4}\)

=>\(2x=-\dfrac{1}{4}\)

=>\(x=-\dfrac{1}{4}:2=-\dfrac{1}{8}\)

b: ĐKXĐ: x>=0

\(\left(6-3\sqrt{x}\right)\left(\left|x\right|-7\right)=0\)

=>\(\left\{{}\begin{matrix}6-3\sqrt{x}=0\\\left|x\right|-7=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}3\sqrt{x}=6\\\left|x\right|=7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x}=2\\\left[{}\begin{matrix}x=7\left(nhận\right)\\x=-7\left(loại\right)\end{matrix}\right.\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=7\left(nhận\right)\\x=4\left(nhận\right)\end{matrix}\right.\)

4 tháng 12 2023

bài nào cũng thấy Phước Thịnh :)

27 tháng 9 2021

a) \(\Leftrightarrow2\left|3x-1\right|=\dfrac{4}{5}\)

\(\Leftrightarrow\left|3x-1\right|=\dfrac{2}{5}\)

\(\Leftrightarrow\left[{}\begin{matrix}3x-1=\dfrac{2}{5}\\3x-1=-\dfrac{2}{5}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{15}\\x=\dfrac{1}{5}\end{matrix}\right.\)

b)TH1:  \(x\ge3\)

\(\Leftrightarrow x+5+x-3=9\Leftrightarrow2x=7\Leftrightarrow x=\dfrac{7}{2}\left(tm\right)\)

TH2: \(-5\le x< 3\)

\(\Leftrightarrow x+5-x+3=9\Leftrightarrow8=9\left(VLý\right)\)

TH3: \(x< -5\)

\(\Leftrightarrow-x-5-x+3=9\Leftrightarrow2x=-11\Leftrightarrow x=-\dfrac{11}{2}\left(tm\right)\)

27 tháng 9 2021

\(a,2.|3x-1|-\dfrac{3}{4}=\dfrac{1}{20}\)

\(2.|3x-1|=\dfrac{1}{20}+\dfrac{3}{4}\)

\(2.|3x-1|=\dfrac{4}{5}\)

\(|3x-1|=\dfrac{4}{5}:2\)

\(|3x-1|=\dfrac{2}{5}\)

\(\Rightarrow3x-1=\pm\dfrac{2}{5}\)

\(3x-1=\dfrac{2}{5}\)

\(3x=\dfrac{2}{5}+1\)

\(3x=\dfrac{7}{5}\)

\(x=\dfrac{7}{5}:3\)

\(x=\dfrac{7}{15}\)

\(3x-1=-\dfrac{2}{5}\)

\(3x=-\dfrac{2}{5}+1\)

\(3x=\dfrac{3}{5}\)

\(x=\dfrac{3}{5}:3\)

\(x=\dfrac{1}{5}\)

`#3107.101107`

a)

`-2/3(x + 1) = 1/6 - x`

`=> -2/3x - 2/3 = 1/6 - x`

`=> -2/3x + x = 1/6 + 2/3`

`=> 1/3x = 5/6`

`=> x = 5/6 \div 1/3`

`=> x =5/2`

Vậy, `x = 5/2`

b)

`3(x + 1/3) - 1/2(x + 2) = 5/2x - 1`

`=> 3x + 1 - 1/2x - 1 = 5/2x - 1`

`=> 3x - 1/2x - 5/2x = -1`

`=> 0x = -1` (vô lý)

Vậy, `x` không có giá trị thỏa mãn.

a: \(\Leftrightarrow-\dfrac{2}{3}x-\dfrac{2}{3}=\dfrac{1}{6}-x\)

=>\(\dfrac{1}{3}x=\dfrac{1}{6}+\dfrac{2}{3}=\dfrac{5}{6}\)

=>\(x=\dfrac{5}{6}\cdot3=\dfrac{5}{2}\)

b: \(\Leftrightarrow3x+1-\dfrac{1}{2}x-1=\dfrac{5}{2}x-1\)

=>\(\dfrac{5}{2}x=\dfrac{5}{2}x-1\)

=>0=-1(vô lý)