Câu g đề bài lỗi nhé e

a,\(\left(x-\frac{2}{3}\right),\left(x+\frac{1}{1}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\frac{2}{3}\\x+\frac{1}{4}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{2}{3}\\x=\frac{-1}{4}\end{matrix}\right.\)

b,\(\left(x-\frac{2}{3}\right)\left(2x-\frac{3}{4}\right)=\left(3x+\frac{1}{2}\right)\left(x+\frac{2}{3}\right)\)

\(\Leftrightarrow2x^2-\frac{3}{4}x-\frac{4}{3}x+\frac{1}{2}=3x^2+2x+\frac{1}{2}x+\frac{1}{3}\)

\(\Leftrightarrow2x^2-\frac{25}{12}x+\frac{1}{2}=3x^2+\frac{5}{2}x+\frac{1}{3}\)

\(\Leftrightarrow24x^2-25x+6=36x^2+30x+4\)

\(\Leftrightarrow24x^2-25x+6-36x^2-30x-4=0\)

\(\Leftrightarrow-12x^2-55x+2=0\)

\(\Leftrightarrow12x^2+55x-2=0\)

\(\Leftrightarrow x=\frac{-55\pm\sqrt{55^2-4.12\left(-2\right)}}{2.12}\)

\(\Leftrightarrow\frac{-55\pm\sqrt{3025+96}}{24}\)

\(\Leftrightarrow\frac{-55\pm\sqrt{3121}}{24}\)

\(\Leftrightarrow\frac{-55+\sqrt{3121}}{24}\)

\(\Leftrightarrow\left[{}\begin{matrix}\frac{-55+\sqrt{3121}}{24}\\\frac{-55-\sqrt{3121}}{24}\end{matrix}\right.\)

\(A=\frac{99}{100}-\left(\frac{1}{1.2}+\frac{1}{2.3}+..+\frac{1}{99.100}\right)\)

\(A=\frac{99}{100}-\left(1-\frac{1}{100}\right)\)

\(A=\frac{99}{100}-\frac{99}{100}\)

\(A=\frac{99-99}{100}=0\)

Bài 2 

\(\left(3x+5\right).\left(2x-4\right)=0\)

\(TH1:3x+5=0\)

\(3x=-5\)

\(x=-\frac{5}{3}\)

\(TH2:2x-4=0\)

\(2x=4\)

\(x=2\)

\(\left(x^2-1\right).\left(x+3\right)=0\)

\(\Rightarrow x^2-1=0\)

\(x^2=1\)

\(\Rightarrow x=1\)

\(x+3=0\)

\(x=-3\)

\(5x^2-\frac{1}{2}x=0\)

\(\Rightarrow5x^2-\frac{x}{2}=0\)

\(\Rightarrow5x^2=\frac{5x^2}{1}=\frac{5x^2.2}{2}\)

\(10x^2-x=x.\left(10x-1\right)\)

\(\frac{x.\left(10x-1\right)}{2}=0\)

\(\frac{x.\left(10x-1\right)}{2}.2=0.2\)

\(10x-1=0\)

\(x=\frac{1}{10}=0.100\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{1}{10}=0.100\\x=0\end{cases}}\)

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

\(\frac{x}{4}=\frac{3}{4}+\frac{1}{2}\)

\(\frac{x}{4}=\frac{5}{4}\)

\(\Rightarrow x=5\)

\(\frac{1}{8}+\frac{7}{8}:x=\frac{3}{4}\)

\(\frac{7}{8}:x=\frac{3}{4}-\frac{1}{8}\)

\(x=\frac{7}{8}:\frac{5}{8}\)

\(x=\frac{56}{40}=\frac{28}{20}=\frac{14}{10}=\frac{7}{5}\)

Câu a:

2.(3\(x\) - \(\frac12\)) - 2\(x\) = \(\frac12\).(2\(x\) - 3)

6\(x\) - 1 - 2\(x\) = \(x\) - \(\frac32\)

6\(x\) - 2\(x\) - \(x\) = 1 - \(\frac32\)

4\(x\) - \(x\) = - \(\frac12\)

3\(x\) = - \(\frac12\)

\(x\) = - \(\frac12\) : 3

\(x=-\frac16\)

Vậy \(x=-\frac16\)

Câu b:

(2\(x\) - \(\frac35\))\(^2\) = \(\frac{4}{25}\)

(2\(x-\frac35\))\(^2\) = \(\left(\frac{2}{25}\right)\)\(^2\)

2\(x\) - \(\frac35\) = \(\frac25\) hoặc 2\(x\) - \(\frac35\) = - \(\frac25\)

TH: 2\(x\) - \(\frac35\) = \(\frac25\)

2\(x\) = \(\frac25+\frac35\)

2\(x\) = 1

\(x=\frac12\)

2\(x\) - \(\frac35\) = - \(\frac25\)

2\(x\) = - \(\frac25\) + \(\frac35\)

2\(x\) = \(\frac15\)

\(x\) = \(\frac{13}{25}\) : 2

\(x\) = \(\frac15\)

Vậy \(x\) ∈ {1/5; 1/2}

co ghi dau ma biet

mk ko chép lại đề nhé bn

b, 

=>\(\left|x-\frac{1}{3}\right|+\frac{4}{5}=\left|-\frac{14}{5}\right|\)

=>\(\left|x-\frac{1}{3}\right|+\frac{4}{5}=\frac{14}{5}\) \(\Rightarrow\left|x-\frac{1}{3}\right|=2\)

\(\Rightarrow\orbr{\begin{cases}x-\frac{1}{3}=-2\\x-\frac{1}{3}=2\end{cases}\Rightarrow\orbr{\begin{cases}x=-\frac{5}{3}\\x=\frac{7}{3}\end{cases}}}\)

c,\(\Rightarrow\frac{x-1}{2013}+\frac{x-2}{2012}-\frac{x-3}{2011}-\frac{x-4}{2010}=0\)

=> \(\frac{x-1}{2013}-1+\frac{x-2}{2012}-1-\left(\frac{x-3}{2011}-1+\frac{x-4}{2010}-1\right)=0\)

=>\(\frac{x-2014}{2013}+\frac{x-2014}{2012}-\frac{x-2014}{2011}-\frac{x-2014}{2010}=0\)

=.\(\left(x-2014\right)\left(\frac{1}{2013}+\frac{1}{2012}-\frac{1}{2011}-\frac{1}{2010}\right)=0\)

Do \(\frac{1}{2013}+\frac{1}{2012}-\frac{1}{2011}-\frac{1}{2010}\ne0\)=> x-2014=0

=> x=2014

d,\(\left(x-7\right)^{x-1}-\left(x-7\right)^{x+11}=0\)

=>\(\left(x-7\right)^{x-1}.\left[1-\left(x-7\right)^{x+12}\right]=0\)

=> \(\orbr{\begin{cases}\left(x-7\right)^{x-1}=0\\1-\left(x-7\right)^{x+12}=0\end{cases}}\)

=> \(\orbr{\begin{cases}x-7=0\\\left(x-7\right)^{x+12}=0\end{cases}}\)

=>x=7 hoặc x-7=1 hoặc x+12=0

=> x=7 hoặc x=8 hoặc x=-12

Vậy x=7, x=8, x=-12

k,3x+x²=0

=> x(3+x)=0

=>\(\orbr{\begin{cases}x=0\\3+x=0\end{cases}}\)

=>\(\orbr{\begin{cases}x=0\\x=-3\end{cases}}\)

m, x²-2x-3(x-2)=0

=> x(x-2)-3(x-2)=0

=> (x-3)(x-2)=0

=>\(\orbr{\begin{cases}x-3=0\\x-2=0\end{cases}}\)

=> \(\orbr{\begin{cases}x=3\\x=2\end{cases}}\)

*****Chúc bạn học giỏi*****

Câu d:

-1\(\frac23\) - (|2\(x\)| + \(\frac56\)) = - 2

-\(\frac53\) - |2\(x\)| - \(\frac56\) = - 2

|2\(x\)| = - \(\frac53\) - \(\frac56\) + 2

|2\(x\)| = - \(\frac52\) + 2

|2\(x\)| = - \(\frac12\) (vô lí vì trị tuyệt đối của một số luôn là một số không âm)

Không có giá trị nào của x thỏa mãn đề bài.

x ∈ ∅

Câu a:

|\(x\) - 3| = \(x\) + 4

Vì |\(x\) - 3| ≥ 0 ∀ \(x\) nên \(x\) + 4 ≥ 0 ⇒ \(x\) ≥ - 4

Với -4 ≤ \(x\) ≤ 3 ta có:

-\(x\) + 3 = \(x\) + 4

\(x\) + \(x\) = -4 + 3

2\(x\) = -1

\(x=\frac{-1}{2}\)

Với x > 3 ta có:

x - 3 = x + 4

x - x = 3 + 4

0 = 7 (vô lí)

Vậy x = -1/2 là nghiện duy nhất của phương trình.

Vậy \(x\) = -1/2

làm hộ mình cái để mai nộp thầy,ai nhanh và đúng thì mình k cho nha

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

\(-\frac{7}{2}+\frac{1}{3}.\left(x-1\right)=-\frac{4}{3}:\frac{7}{3}\)

\(-\frac{7}{2}+\frac{1}{3}.\left(x-1\right)=-\frac{4}{7}\)

\(\frac{1}{3}.\left(x-1\right)=-\frac{4}{7}-\frac{-7}{2}\)

\(\frac{1}{3}.\left(x-1\right)=\frac{41}{14}\)

\(\Rightarrow x-1=\frac{41}{14}:\frac{1}{3}\)

\(\Rightarrow x-1=\frac{123}{14}\)

\(\Rightarrow x=\frac{123}{14}+1\)

\(\Rightarrow x=\frac{137}{14}\)

a, \(-\frac{5}{7}-\left(\frac{1}{2}-x\right)=-\frac{11}{4}\)

\(\frac{1}{2}-x=\frac{57}{28}\)

\(x=-\frac{43}{28}\)

b, \(\left(2x-1\right)^2-5=20\)

\(\Rightarrow\left(2x-1\right)^2=25\)

\(\Rightarrow2x-1=\pm5\)

\(\Rightarrow\left[{}\begin{matrix}2x-1=5\\2x-1=-5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=6\\2x=-4\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)

Câd

\(\frac{x-6}{4}=\frac{4}{x-6}\)

(\(x-6\))(\(x-6\)) =4.4

(\(x-6\))\(^2\) = 4\(^2\)

\(x-6=-4\) hoặc \(x\) - 6 = 4

\(x-6\) = -4

\(x=-4+6\)

\(x=2\)

\(x-6=4\)

\(x=4+6\)

\(x=10\)

Vậy \(x\) ∈ {2; 10}

b, \(\left(2x-1\right)^2-5=20\)

\(\Rightarrow\left(2x-1\right)^2=25\)

\(\Rightarrow\left(2x-1\right)^2=5^2\)

\(\Rightarrow\left[{}\begin{matrix}2x-1=6\\2x-1=-6\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}2x=7\\2x=-5\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\frac{7}{2}\\x=-\frac{5}{2}\end{matrix}\right.\)

Vậy ...

a) \(-\frac{5}{7}-\left(\frac{1}{2}-x\right)=\frac{-11}{4}\)

\(\Rightarrow\left(\frac{1}{2}-x\right)=\left(-\frac{5}{7}\right)+\frac{11}{4}\)

\(\Rightarrow\frac{1}{2}-x=\frac{57}{28}\)

\(\Rightarrow x=\frac{1}{2}-\frac{57}{28}\)

\(\Rightarrow x=-\frac{43}{28}\)

Vậy \(x=-\frac{43}{28}.\)

b) \(\left(2x-1\right)^2-5=20\)

\(\Rightarrow\left(2x-1\right)^2=20+5\)

\(\Rightarrow\left(2x-1\right)^2=25\)

\(\Rightarrow2x-1=\pm5\)

\(\Rightarrow\left[{}\begin{matrix}2x-1=5\\2x-1=-5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=5+1=6\\2x=\left(-5\right)+1=-4\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=6:2\\x=\left(-4\right):2\end{matrix}\right.\)

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

Vậy \(x\in\left\{3;-2\right\}.\)

d) \(\frac{x-6}{4}=\frac{4}{x-6}\)

\(\Rightarrow\left(x-6\right).\left(x-6\right)=4.4\)

\(\Rightarrow\left(x-6\right).\left(x-6\right)=16\)

\(\Rightarrow\left(x-6\right)^2=16\)

\(\Rightarrow x-6=\pm4\)

\(\Rightarrow\left[{}\begin{matrix}x-6=4\\x-6=-4\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=4+6\\x=\left(-4\right)+6\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=10\\x=2\end{matrix}\right.\)

Vậy \(x\in\left\{10;2\right\}.\)

Chúc bạn học tốt!