\(B=x+\dfrac{0,2-0,375+\dfrac{5}{11}}{-0,3+\dfrac{9}{16}-\dfrac{15}{22}}\)

\(=x+\dfrac{\dfrac{1}{5}-\dfrac{3}{8}+\dfrac{5}{11}}{-\left(\dfrac{3}{10}-\dfrac{9}{16}+\dfrac{15}{22}\right)}\)

\(=x+\dfrac{\dfrac{1}{5}-\dfrac{3}{8}+\dfrac{5}{11}}{-\dfrac{3}{2}\left(\dfrac{1}{5}-\dfrac{3}{8}+\dfrac{5}{11}\right)}\)

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

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

Với \(x=-\dfrac{1}{3}\), ta được:

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

B = -1

7) 5x=4y ⇒\(\dfrac{x}{4}=\dfrac{y}{5}\)

Nhân cả hai vế với \(\dfrac{x}{4}\), ta có: \(\left(\dfrac{x}{4}\right)^2=\dfrac{x}{4}.\dfrac{y}{5}=\dfrac{xy}{20}=\dfrac{20}{20}=1\)

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

⇒ \(\left[{}\begin{matrix}y=5\\y=-5\end{matrix}\right.\)

4) áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\dfrac{x}{0,5}=\dfrac{y}{0,3}=\dfrac{z}{0,2}=\dfrac{z-y+x}{0,2-0,3+0,5}=\dfrac{1}{\dfrac{2}{5}}=\dfrac{5}{2}\)

\(\dfrac{x}{0,5}=\dfrac{5}{2}\Rightarrow x=\dfrac{5}{4}\)

\(\dfrac{y}{0,3}=\dfrac{5}{2}\Rightarrow y=\dfrac{3}{4}\)

\(\dfrac{z}{0,2}=\dfrac{5}{2}\Rightarrow z=\dfrac{1}{2}\)

6) áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\dfrac{x+11}{13}=\dfrac{y+12}{14}=\dfrac{z+13}{15}=\dfrac{x+11+y+12+z+13}{13+14+15}=\dfrac{42}{42}=1\)

\(\dfrac{x+11}{13}=1\Rightarrow x=2\)

\(\dfrac{y+12}{13}=1\Rightarrow y=1\)

\(\dfrac{z+13}{15}=1\Rightarrow z=2\)

7) \(5x=4y\Rightarrow\dfrac{x}{4}=\dfrac{y}{5}=k\)

\(\Rightarrow x=4k,y=5k\)

\(x.y=20\\ \Rightarrow4k.5k=20\\ \Rightarrow20k^2=20\\ \Rightarrow k^2=1\\ \Rightarrow\left[{}\begin{matrix}k=-1\\k=1\end{matrix}\right.\)

\(x=4k\Rightarrow\left[{}\begin{matrix}x=-4\\x=4\end{matrix}\right.\)

\(y=5k\Rightarrow\left[{}\begin{matrix}y=-5\\y=5\end{matrix}\right.\)

Vậy \(\left(x,y\right)=\left\{\left(-4;-5\right);\left(4;5\right)\right\}\)

\(B=x+\dfrac{0,2-0,375+\dfrac{5}{11}}{-0,3+\dfrac{9}{16}-\dfrac{15}{22}}\)

\(=x+\dfrac{0,2-0,375+\dfrac{5}{11}}{-0,3+\dfrac{9}{16}-\dfrac{15}{22}}\)

\(=x+\dfrac{-\dfrac{7}{40}+\dfrac{5}{11}}{-\dfrac{369}{880}}\)

\(=x+\dfrac{\dfrac{123}{440}}{\dfrac{369}{880}}\)

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

Thay \(x=-\dfrac{1}{3}\) vào biểu thức B.

Ta có: \(-\dfrac{1}{3}-\dfrac{2}{3}=-\dfrac{3}{3}=-1\)

Vậy giá trị biểu thức B tại \(x=-\dfrac{1}{3}\) là -1.

a, \(\dfrac{3}{4}x=-\dfrac{9}{8}\)
x= \(-\dfrac{3}{2}\)
b, |x| + 0,25= 5,25
|x | = 5
=> x\(\in\){ +- 5}
Ko chắc đúng, kiểm tra trc khi làm

\(\dfrac{2x-1}{3}=\dfrac{-5}{0.6}\)

\(\Leftrightarrow2x-1=-25\)

hay x=-12

\(A=x+152=\dfrac{1}{3}+152=\dfrac{457}{3}\)

a)-12.(8)

b) -0.6

a) \(-0,6x-\dfrac{7}{3}=5,4\Leftrightarrow-\dfrac{3}{5}x=5,4+\dfrac{7}{3}\Leftrightarrow x=\dfrac{116}{15}.\left(-\dfrac{5}{3}\right)=-\dfrac{116}{9}\).

b) \(2,8:\left(\dfrac{1}{5}-3x\right)=1\dfrac{2}{5}\Leftrightarrow\dfrac{1}{5}-3x=2,8:\dfrac{7}{5}\Leftrightarrow-3x=2-\dfrac{1}{5}\Leftrightarrow x=\dfrac{9}{5}:\left(-3\right)=-\dfrac{3}{5}\).

hay x = -12

Tìm x biết : \(\dfrac{2x-1}{3}=\dfrac{-5}{0,6}\)

\(2x-1=3.\left(\dfrac{-5}{0,6}\right)\)

\(2x-1=3.\left(\dfrac{-50}{6}\right)\)

\(2x-1=3\left(\dfrac{-25}{3}\right)\)

\(2x-1=-25\)

\(2x=-25+1\)

\(2x=-24\)

\(x=\dfrac{-24}{2}\)

\(x=-12\)

x = -12 nha

1: Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{x}{0,3}=\dfrac{y}{0.2}=\dfrac{z}{0.1}=\dfrac{x-y}{0.3-0.2}=\dfrac{1}{0.1}=10\)

Do đó: x=3; y=2; z=1