Ta có 4x³ - 3  = 29

=> 4x³ = 32

=> x³ = 8

=> x³ = 2³

=> x = 2

Khi x = 2 => \(\frac{25-y}{16}=\frac{z+49}{25}=\frac{8}{9}\)(1)

=> \(\frac{25-y}{16}=\frac{8}{9}\)

=> 9(25 - y) = 128

=> y = 97/9

Từ (1) => \(\frac{z+49}{25}=\frac{8}{9}\)

=> 9(z + 49) = 25.8

=> z = -241/9 

Ta có: 4x³ - 3 = 29 <=> 4x³ = 32 <=> x³ = 8 <=>x = 2

Do đó: \(\frac{2+6}{9}=\frac{25-y}{16}=\frac{z+49}{25}\)

=> \(\hept{\begin{cases}\frac{25-y}{16}=\frac{8}{9}\\\frac{z+49}{25}=\frac{8}{9}\end{cases}}\) => \(\hept{\begin{cases}225-9y=128\\9z+441=200\end{cases}}\) => \(\hept{\begin{cases}y=\frac{97}{9}\\z=-\frac{421}{9}\end{cases}}\)

bn chụp lại đi chụp thẳng ý

b) \(\frac{10-x}{100}+\frac{20-x}{110}+\frac{30-x}{120}=3\)

\(\Leftrightarrow\frac{10-x}{100}-1+\frac{20-x}{110}-1+\frac{30-x}{120}-1=0\)

\(\Leftrightarrow\frac{-x-90}{100}+\frac{-x-90}{110}+\frac{-x-90}{120}=0\)

\(\Leftrightarrow-x-90\left(\frac{1}{100}+\frac{1}{110}+\frac{1}{120}\right)=0\)

\(\Rightarrow-x-90=0\)

Vì \(\frac{1}{100}+\frac{1}{110}+\frac{1}{120}\ne0\)

\(-x-90=0\)

\(-x=0+90\)

\(-x=90\)

\(\Rightarrow x=-90\)

\(\frac{x+1}{2}+\frac{x+5}{3}+\frac{x+11}{4}+\frac{x+19}{5}=10\)

\(\Rightarrow\frac{x+1}{2}-1+\frac{x+5}{3}-2+\frac{x+11}{4}-3+\frac{x+19}{5}-4=0\)

\(\Rightarrow\frac{x-1}{2}+\frac{x-1}{3}+\frac{x-1}{4}+\frac{x-1}{5}=0\)

\(\Rightarrow\left(x-1\right)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}\right)=0\)

\(\Rightarrow x-1=0\)Vì \(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}\ne0\)

\(\Rightarrow x=1\)

Ta có :

\(A=\frac{\frac{3}{7}-\frac{3}{17}+\frac{3}{37}}{\frac{5}{7}-\frac{5}{17}+\frac{5}{37}}+\frac{\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}}{\frac{7}{2}-\frac{7}{3}+\frac{7}{4}-\frac{7}{5}}\)

\(\Rightarrow A=\frac{3\left(\frac{1}{7}-\frac{1}{17}+\frac{1}{37}\right)}{5\left(\frac{1}{7}-\frac{1}{17}+\frac{1}{37}\right)}+\frac{\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}}{7\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}\right)}\)

\(\Rightarrow A=\frac{3}{5}+\frac{1}{7}=\frac{21}{35}+\frac{5}{35}=\frac{26}{35}\)

b) Ta có \(\frac{7}{3.6}+\frac{7}{6.9}+\frac{7}{9.12}+...+\frac{7}{66.69}=\frac{7}{3}.\left(\frac{3}{3.6}+\frac{3}{6.9}+\frac{3}{9.12}+...+\frac{3}{66.69}\right)\)

\(=\frac{7}{3}\left(\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+\frac{1}{9}-\frac{1}{12}+...+\frac{1}{66}-\frac{1}{69}\right)\)

\(=\frac{7}{3}\left(\frac{1}{3}-\frac{1}{69}\right)=\frac{7}{3}.\frac{22}{69}=\frac{154}{207}\)

Lại có \(\frac{5}{1.3}+\frac{5}{3.5}+\frac{5}{5.7}+...+\frac{5}{67.69}=\frac{5}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{67.69}\right)\)

\(=\frac{5}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{67}-\frac{1}{69}\right)\)

\(=\frac{5}{2}\left(1-\frac{1}{69}\right)=\frac{5}{2}.\frac{68}{69}=\frac{170}{69}\)

Khi đó B = \(\frac{154}{207}:\frac{170}{69}=\frac{154.69}{207.170}=\frac{77}{255}\)

ai làm ny mk nha

Rảnh zậy

\(\frac{x+1}{2020}+\frac{x+2}{2019}=\frac{x+3}{2018}+\frac{x}{2021}\)

\(\Leftrightarrow\frac{x+1}{2020}+1+\frac{x+2}{2019}+1=\frac{x+3}{2018}+1+\frac{x}{2021}+1\)

\(\Leftrightarrow\frac{x+2021}{2020}+\frac{x+2021}{2019}=\frac{x+2021}{2018}+\frac{x+2021}{2021}\)

\(\Leftrightarrow x+2021=0\)

\(\Leftrightarrow x=-2021\)