:v lớp 10

D

mik làm ngày 2 nhé.

Bài 1:

a)x=3

b)x=2

c)x=2

d)x=0

bạn kết bạn với mik để mik làm nốt cho nhé.

\(\Rightarrow\dfrac{5}{4}-\dfrac{1}{4}x=\dfrac{3}{10}x-\dfrac{2}{5}\)

\(\Rightarrow\dfrac{5}{4}+\dfrac{2}{5}=\dfrac{3}{10}x-\dfrac{1}{4}x\)

\(\Rightarrow\dfrac{33}{20}=\dfrac{11}{20}x\)

\(\Rightarrow x=\dfrac{33}{20}\div\dfrac{11}{20}\)

\(\Rightarrow x=3\)

\(1\dfrac{1}{4}-x\dfrac{1}{4}=x\cdot30\%\cdot\dfrac{2}{5}\)

\(\Leftrightarrow\dfrac{5}{4}-x\dfrac{1}{4}=x\cdot\dfrac{3}{10}-\dfrac{2}{5}\)

\(\Leftrightarrow\dfrac{5}{4}-\dfrac{1}{4}x=\dfrac{3}{10}x-\dfrac{2}{5}\)

\(\Leftrightarrow25-5x=6x-8\)

\(\Leftrightarrow-5x-6x=-8-25\)

\(\Leftrightarrow-11x=-33\)

\(\Leftrightarrow x=3\)

Vậy x = 3

sao thi muộn vậy bn ! mk thi xong lâu rùi

i) \(5\dfrac{8}{17}:x+\left(-\dfrac{4}{17}\right):x+3\dfrac{1}{7}:17\dfrac{1}{3}=\dfrac{4}{11}\)

\(\Rightarrow\dfrac{93}{17}:x-\dfrac{4}{17}:x+\dfrac{33}{182}=\dfrac{4}{11}\)

\(\Rightarrow\left(\dfrac{93}{17}-\dfrac{4}{17}\right):x=\dfrac{4}{11}-\dfrac{33}{182}\)

\(\Rightarrow\dfrac{89}{17}:x=\dfrac{365}{2002}\)

\(\Rightarrow x=\dfrac{89}{17}:\dfrac{365}{2002}=\dfrac{178178}{6205}\)

j) \(\dfrac{17}{2}-\left|2x-\dfrac{3}{4}\right|=-\dfrac{7}{4}\)

\(\Rightarrow\left|2x-\dfrac{3}{4}\right|=\dfrac{17}{2}-\left(-\dfrac{7}{4}\right)=\dfrac{41}{4}\)

\(\Rightarrow\left[{}\begin{matrix}2x-\dfrac{3}{4}=\dfrac{41}{4}\\2x-\dfrac{3}{4}=-\dfrac{41}{4}\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}2x=11\Rightarrow x=\dfrac{11}{2}\\2x=-\dfrac{19}{2}\Rightarrow x=-\dfrac{19}{4}\end{matrix}\right.\)

k) \(\left(x+\dfrac{1}{5}\right)^2+\dfrac{17}{25}=\dfrac{26}{25}\)

\(\Rightarrow\left(x+\dfrac{1}{5}\right)^2=\dfrac{26}{25}-\dfrac{17}{25}=\dfrac{9}{25}=\left(\dfrac{3}{5}\right)^2\)\(=\left(-\dfrac{3}{5}\right)^2\)

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

l) \(-1\dfrac{5}{27}-\left(3x-\dfrac{7}{9}\right)^3=-\dfrac{24}{27}\)

\(\Rightarrow\left(3x-\dfrac{7}{9}\right)^3=\dfrac{-32}{27}-\left(-\dfrac{24}{27}\right)=-\dfrac{8}{27}=\left(-\dfrac{2}{3}\right)^3\)

\(\Rightarrow3x-\dfrac{7}{9}=-\dfrac{2}{3}\)

\(\Rightarrow3x=-\dfrac{2}{3}+\dfrac{7}{9}=\dfrac{1}{9}\)

\(\Rightarrow x=\dfrac{1}{27}\)

j, \(\dfrac{17}{2}-\left|2x-\dfrac{3}{4}\right|=\dfrac{-7}{4}\)

\(\Rightarrow-\left|2x-\dfrac{3}{4}\right|=\dfrac{-7}{4}-\dfrac{17}{2}\)

\(\Rightarrow-\left|2x-\dfrac{3}{4}\right|=\dfrac{-41}{4}\)

\(\Rightarrow\left|2x-\dfrac{3}{4}\right|=\dfrac{41}{4}\)

\(\Rightarrow\left[{}\begin{matrix}2x-\dfrac{3}{4}=\dfrac{41}{4}\\2x-\dfrac{3}{4}=\dfrac{-41}{4}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{11}{2}\\x=\dfrac{-19}{4}\end{matrix}\right.\)

k, \(\left(x+\dfrac{1}{5}\right)^2+\dfrac{17}{25}=\dfrac{26}{25}\)

\(\Rightarrow\left(x+\dfrac{1}{5}\right)^2=\dfrac{9}{25}\)

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

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

l, \(-1\dfrac{5}{27}-\left(3x-\dfrac{7}{9}\right)^3=\dfrac{-24}{27}\)

\(\Rightarrow-\left(3x-\dfrac{7}{9}\right)^3=\dfrac{-19}{27}\)

\(\Rightarrow\left(3x-\dfrac{7}{9}\right)^3=\dfrac{19}{27}\)

\(\Rightarrow3x-\dfrac{7}{9}=\dfrac{\sqrt[3]{19}}{3}\)

\(\Rightarrow3x=\dfrac{\sqrt[3]{19}}{3}+\dfrac{7}{19}\)

\(\Rightarrow...\)

a) 4126

b) 615

c) 927

b) 615

c) 927

Ta có: \(\dfrac{1}{2}\cdot y+\dfrac{2}{3}\cdot y=\dfrac{7}{6}\Rightarrow y\left(\dfrac{1}{2}+\dfrac{2}{3}\right)=\dfrac{7}{6}\Rightarrow\dfrac{7}{6}y=\dfrac{7}{6}\Rightarrow y=\dfrac{7}{6}:\dfrac{7}{6}=1\)

Vậy \(D=\left\{1\right\}\)

1

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