Ta có : \(\frac{x}{3}=\frac{y}{4}\Rightarrow\frac{x}{9}=\frac{y}{12}\left(1\right)\)

           \(\frac{y}{6}=\frac{z}{5}\Rightarrow\frac{y}{12}=\frac{z}{10}\left(2\right)\)

Từ (1) và (2) => \(\frac{x}{9}=\frac{y}{12}=\frac{z}{10}\)

Ta có : \(\frac{x}{9}=\frac{y}{12}=\frac{z}{10}=\frac{3x}{27}=\frac{2y}{24}=\frac{5z}{50}=\frac{3x-2y+5z}{27-24+50}=\frac{86}{53}\) (đề sai)

b) Đặt : k = \(\frac{x}{5}=\frac{y}{7}\)

=> k² \(=\frac{x}{5}.\frac{y}{7}=\frac{xy}{35}=\frac{140}{35}=4\)

=> k = -2;2

+ k = 2 thì \(\frac{x}{5}=2\Rightarrow x=10\)

                 \(\frac{z}{7}=2\Rightarrow z=14\)

+ k = -2 thì \(\frac{x}{5}=2\Rightarrow x=-10\)

                 \(\frac{z}{7}=2\Rightarrow z=-14\)

Vậy................................

\(C=\left|\frac{4}{9}-\left(\frac{\sqrt{2}}{2}\right)^2\right|+\left|0,\left(4\right)+\frac{\frac{1}{3}-\frac{2}{5}-\frac{3}{7}}{\frac{2}{3}-\frac{4}{5}-\frac{6}{7}}\right|\)

\(=\left|\frac{4}{9}-\frac{1}{2}\right|+\left|\frac{4}{9}+\frac{\frac{1}{3}-\frac{2}{5}-\frac{3}{7}}{2\left(\frac{1}{3}-\frac{2}{5}-\frac{3}{7}\right)}\right|\)

\(=\frac{1}{18}+\left|\frac{4}{9}+\frac{1}{2}\right|\)

\(=\frac{1}{18}+\frac{17}{18}=1\)

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

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

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

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

Bài 2 

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

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

\(3x=-5\)

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

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

\(2x=4\)

\(x=2\)

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

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

\(x^2=1\)

\(\Rightarrow x=1\)

\(x+3=0\)

\(x=-3\)

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

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

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

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

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

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

\(10x-1=0\)

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

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

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

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

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

\(\Rightarrow x=5\)

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

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

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

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

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

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

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

\(=0\)

Bài 2:

a) \(x:\left(\frac{2}{9}-\frac{1}{5}\right)=\frac{8}{16}\)

\(\Leftrightarrow x:\frac{1}{45}=\frac{1}{2}\)

\(\Leftrightarrow x=\frac{1}{2}:\frac{1}{45}=\frac{45}{2}\)

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

\(\)\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\2x+3=0\end{matrix}\right.\)

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

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

c) \(\frac{4-3x}{2x+5}=0\Leftrightarrow4-3x=0\)

\(\Leftrightarrow3x=4\Rightarrow x=\frac{4}{3}\)

d) \(\left(x-2\right).\left(x+\frac{2}{3}\right)\ge0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-2>0\\x+\frac{3}{2}>0\end{matrix}\right.\\\left\{{}\begin{matrix}x-2< 0\\x+\frac{3}{2}< 0\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>2\\x>-\frac{3}{2}\end{matrix}\right.\\\left\{{}\begin{matrix}x< 2\\x< -\frac{3}{2}\end{matrix}\right.\end{matrix}\right.\)

Bài 2:

a) \(x:\left(\frac{2}{9}-\frac{1}{5}\right)=\frac{8}{16}\)

=> \(x:\frac{1}{45}=\frac{1}{2}\)

=> \(x=\frac{1}{2}.\frac{1}{45}\)

=> \(x=\frac{1}{90}\)

Vậy \(x=\frac{1}{90}.\)

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

=> \(\left\{{}\begin{matrix}2x-1=0\\2x+3=0\end{matrix}\right.\) => \(\left\{{}\begin{matrix}2x=0+1=1\\2x=0-3=-3\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x=1:2\\x=\left(-3\right):2\end{matrix}\right.\)

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

Vậy \(x\in\left\{\frac{1}{2};-\frac{3}{2}\right\}.\)

Mình chỉ làm được thế thôi nhé, mong bạn thông cảm.

Chúc bạn học tốt!

1. \(\frac{7x-1}{6}+2x=\frac{16-x}{5}\)

\(\Leftrightarrow5\left(7x-1\right)+60x=6\left(16-x\right)\)

\(\Leftrightarrow35x-5+60x=96-6x\)

\(\Leftrightarrow95x-5=96-6x\)

\(\Leftrightarrow95x+6x=96+5\)

\(\Leftrightarrow101x=101\)

\(\Leftrightarrow x=1\)

2. \(\frac{10x+3}{12}=1+\frac{6+8x}{9}\) 

\(\Leftrightarrow3\left(10x+3\right)=36+4\left(6+8x\right)\)

\(\Leftrightarrow30x+9=36+24+32x\)

\(\Leftrightarrow30x+9=32x+60\)

\(\Leftrightarrow30x-32x=60-9\)

\(\Leftrightarrow-2x=51\)

\(\Leftrightarrow x=-\frac{51}{2}\)

3. \(\frac{8x-3}{4}-\frac{3x-2}{2}=\frac{2x-1}{2}+\frac{x+3}{4}\)

\(\Leftrightarrow8x-3-2\left(3x-2\right)=2\left(2x-1\right)+x+3\)

\(\Leftrightarrow8x-3-6x+4=4x-2+x+3\)

\(\Leftrightarrow2x+1=5x+1\)

\(\Leftrightarrow2x=5x\)

\(\Leftrightarrow x=0\)

4) \(\frac{3\left(3-x\right)}{8}+\frac{2\left(5-x\right)}{3}=\frac{1-x}{2}-2\)

=> \(\frac{9-3x}{8}+\frac{10-2x}{3}=\frac{1-x}{2}-\frac{2}{1}\)

=> \(\frac{3\left(9-3x\right)}{24}+\frac{8\left(10-2x\right)}{24}=\frac{12\left(1-x\right)}{24}-\frac{48}{24}\)

=> \(\frac{27-9x}{24}+\frac{80-16x}{24}=\frac{12-12x}{24}-\frac{48}{24}\)

=> \(\frac{27-9x+80-16x}{24}=\frac{12-12x-48}{24}\)

=> 27 - 9x + 80 - 16x = 12 - 12x - 48

=> 27 - 9x + 80 - 16x - 12 + 12x + 48 = 0

=> (27 + 80 - 12 + 48) + (-9x - 16x + 12x) = 0

=> 143 - 13x = 0

=> 13x = 143

=> x = 11

5) \(\frac{2\left(x-3\right)}{7}+\frac{x-5}{3}-\frac{13x+4}{21}=0\)

=> \(\frac{2x-6}{7}+\frac{x-5}{3}-\frac{13x+4}{21}=0\)

=> \(\frac{3\left(2x-6\right)}{21}+\frac{7\left(x-5\right)}{21}-\frac{13x+4}{21}=0\)

=> \(\frac{6x-18}{21}+\frac{7x-35}{21}-\frac{13x+4}{21}=0\)

=> \(\frac{6x-18+7x-35-13x-4}{21}=0\)

=> 6x - 18 + 7x - 35 - 13x - 4 = 0

=> (6x + 7x - 13x) + (-18 - 35 - 4) = 0

=> -57 = 0(vô nghiệm)

6) \(\frac{6x+5}{2}-\left(2x+\frac{2x+1}{2}\right)=\frac{10x+3}{4}\)

=> \(\frac{6x+5}{2}-\frac{10x+3}{4}=2x+\frac{2x+1}{2}\)

=> \(\frac{2\left(6x+5\right)}{4}-\frac{10x+3}{4}=\frac{8x}{4}+\frac{2\left(2x+1\right)}{4}\)

=> \(\frac{12x+10}{4}-\frac{10x+3}{4}=\frac{8x}{4}+\frac{4x+2}{4}\)

=> \(\frac{12x+10-\left(10x+3\right)}{4}=\frac{8x+4x+2}{4}\)

=> \(\frac{12x+10-10x-3}{4}=\frac{12x+2}{4}\)

=> \(12x+10-10x-3=12x+2\)

=> \(2x+10-3=12x+2\)

=> 2x + 10 - 3 - 12x - 2 = 0

=> (2x - 12x) + (10 - 3 - 2) = 0

=> -10x + 5 = 0

=> -10x = -5

=> x = 1/2

7) \(\frac{2x-1}{5}-\frac{x-2}{3}-\frac{x+7}{15}=0\)

=> \(\frac{3\left(2x-1\right)}{15}-\frac{5\left(x-2\right)}{15}-\frac{x+7}{15}=0\)

=> \(\frac{6x-3}{15}-\frac{5x-10}{15}-\frac{x+7}{15}=0\)

=> \(\frac{6x-3-\left(5x-10\right)-\left(x+7\right)}{15}=0\)

=> 6x - 3 - 5x + 10 - x - 7 = 0

=> (6x - 5x - x) + (-3 + 10 - 7) = 0

=> 0x + 0 = 0

=> 0x = 0

=> x tùy ý

Bài 8 tự làm nhé

\(=\frac{16}{5}.\frac{15}{16}-\left(\frac{3}{4}+\frac{2}{7}\right):\left(\frac{-29}{28}\right)\)

\(=3-\left(\frac{21}{28}+\frac{8}{28}\right):\left(\frac{-29}{28}\right)\)

\(=3-\left(\frac{29}{28}\right).\left(\frac{-28}{29}\right)\)

\(=3-\left(-1\right)\)

\(=4\)

b)   \(=\left(\frac{1}{4}+\frac{25}{2}-\frac{5}{16}\right):\left(12-\frac{7}{12}:\left(\frac{3}{8}-\frac{1}{12}\right)\right)\)

       \(=\left(\frac{4}{16}+\frac{200}{16}-\frac{5}{16}\right):\left(12-\frac{7}{12}:\left(\frac{3.3}{2.3.4}-\frac{2}{2.3.4}\right)\right)\)

     \(=\left(\frac{199}{16}\right):\left(12-\frac{7}{12}:\left(\frac{9}{24}-\frac{2}{24}\right)\right)\)

      \(=\frac{199}{16}:\left(12-\frac{7}{12}.\frac{24}{7}\right)\)

    \(=\frac{199}{16}:\left(12-2\right)\)

\(=\frac{199}{16}:10\)

\(=\frac{199}{160}\)

c)   \(\left(\frac{-3}{5}+\frac{5}{11}\right):\frac{-3}{7}+\left(\frac{-2}{5}+\frac{6}{5}\right):\frac{-3}{7}\)

\(\left(\frac{-33}{55}+\frac{25}{55}\right):\frac{-3}{7}+\left(\frac{4}{5}\right):\frac{-3}{7}\)

\(\left(\frac{-8}{55}\right).\frac{-7}{3}+\frac{4}{5}.\frac{-7}{3}\)

\(\frac{-7}{3}\left(\frac{-8}{55}+\frac{4}{5}\right)\)

\(\frac{-7}{3}.\frac{36}{55}=\frac{-84}{55}\)

giờ mk phải đi ngủ r mai mk làm cho 

nhớ làm giúp mình nhá mai mình phải đi học r :<

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

b, \(\left|\frac{5}{18}-x\right|-\frac{7}{24}=0\)

\(\Leftrightarrow\orbr{\begin{cases}\frac{5}{18}-x=\frac{7}{24}\\\frac{5}{18}-x=-\frac{7}{24}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-\frac{1}{72}\\x=\frac{41}{72}\end{cases}}\)

c, \(\frac{2}{5}-\left|\frac{1}{2}-x\right|=6\Leftrightarrow\left|\frac{1}{2}-x\right|=-\frac{28}{5}\)vô lí 

Vì \(\left|\frac{1}{2}-x\right|\ge0\forall x\)*luôn dương* Mà \(-\frac{28}{5}< 0\)

=> Ko có x thỏa mãn 

\(|x+\frac{1}{3}|=0\)

\(< =>x+\frac{1}{3}=0< =>x=-\frac{1}{3}\)

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

\(< =>\orbr{\begin{cases}x+\frac{3}{4}=\frac{1}{2}\\x+\frac{3}{4}=-\frac{1}{2}\end{cases}}\)

\(< =>\orbr{\begin{cases}x=-\frac{1}{4}\\x=-\frac{5}{4}\end{cases}}\)

\(|\frac{5}{18}-x|-\frac{7}{24}=0\)

\(< =>|\frac{5}{18}-x|=\frac{7}{24}\)

\(< =>\orbr{\begin{cases}\frac{5}{18}-x=\frac{7}{24}\\\frac{5}{18}-x=-\frac{7}{24}\end{cases}}\)

\(< =>\orbr{\begin{cases}x=-\frac{1}{72}\\x=\frac{41}{72}\end{cases}}\)

\(\frac{2}{5}-|\frac{1}{2}-x|=6\)

\(< =>\frac{2}{5}-6=|\frac{1}{2}-x|\)

\(< =>\orbr{\begin{cases}\frac{1}{2}-x=-\frac{28}{5}\\\frac{1}{2}-x=\frac{28}{5}\end{cases}}\)

\(< =>\orbr{\begin{cases}x=\frac{61}{10}\\x=-\frac{51}{10}\end{cases}}\)