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\(\frac{x}{4}=\frac{-15}{y}=\frac{z}{52}=\frac{-1}{2}=\frac{-2}{4}=\frac{-15}{30}=\frac{-26}{52}\Rightarrow\left\{{}\begin{matrix}x=-2\\y=30\\z=-26\end{matrix}\right.\)
\(\frac{x}{-4}=\frac{45}{y}=\frac{8}{-12}=\frac{-2}{3}=\frac{\frac{8}{3}}{-4}\left(\text{loại vì x nguyên}\right)\)
\(a)\frac{x}{8}=\frac{-30}{y}=\frac{-48}{32}\)
Rút gọn : \(\frac{-48}{32}=\frac{(-48):16}{32:16}=\frac{-3}{2}\)
* Ta có : \(\frac{x}{8}=\frac{-3}{2}\)
\(\Rightarrow x\cdot2=-3\cdot8\)
\(\Rightarrow x=\frac{-3\cdot8}{2}=-12\)
* Ta có : \(\frac{-30}{y}=\frac{-3}{2}\)
\(\Rightarrow-30\cdot2=-3\cdot y\)
\(\Rightarrow y=\frac{-30\cdot2}{-3}=20\)
Mấy bài kia làm tương tự
\(4\frac{1}{3}.\left(\frac{1}{6}-\frac{1}{2}\right)\le x\le\frac{2}{3}.\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{4}\right)\)
\(\frac{13}{3}.\frac{-1}{3}\le x\le\frac{2}{3}.\frac{-1}{12}\)
\(\frac{-13}{9}\le x\le\frac{-1}{18}\)
\(\Rightarrow\frac{-26}{18}\le x\le\frac{-1}{18}\)
\(\Rightarrow x\in\left(\frac{-26}{18};...;\frac{-18}{18};...\frac{-1}{18}\right)\)
mà x phải thuộc Z
\(\Rightarrow x=\frac{-18}{18}=-1\)
CHÚC BN HỌC TỐT!
=>(5/17+12/17)+(-20/31-11/31)-4/9<=x/9<=(-3/7-4/7)+(7/15+8/15)+2/3
=>-4/9<=x/9<=6/9
=>-4<=x<=6
hay \(x\in\left\{-4;-3;-2;-1;0;...;6\right\}\)
Bài 1:
\(S=4\left(\dfrac{1}{1\cdot7}+\dfrac{1}{7\cdot13}+...+\dfrac{1}{43\cdot49}\right)\)
\(=\dfrac{4}{6}\left(\dfrac{6}{1\cdot7}+\dfrac{6}{7\cdot13}+...+\dfrac{6}{43\cdot49}\right)\)
\(=\dfrac{2}{3}\left(1-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{13}+...+\dfrac{1}{43}-\dfrac{1}{49}\right)\)
\(=\dfrac{2}{3}\cdot\dfrac{48}{49}=\dfrac{96}{147}=\dfrac{32}{49}\)
Bài 3:
Theo đề, ta có:
\(\dfrac{a}{b}=\dfrac{a+10}{b+10}\)
=>ab+10a=ab+10b
=>10a=10b
=>a/b=1
b) 52-\(|\)x\(|\)=-80
\(|\)x\(|\)=52-(-80)
\(|\)x\(|\)=52+80
\(|\)x\(|\)=132
Vậy x=-132
Ta có:12/9=8/x
<=>12/9=(2/3.8)/(2/3.x)
<=>12/9=12/(2/3x)
<=>9=2/3x
<=>x=2/3.9=6
tương tự y=28
rút gọn=>x=3;y=4=>x^2+y^2=3^2+4^2=9+16=25
phần trên mk thiếu đây là phần còn lại
\(\frac{x}{2}=\frac{3}{9}\)
\(\Leftrightarrow\frac{x}{2}=\frac{1}{3}\)
\(\Leftrightarrow3x=2\)
\(\Leftrightarrow x=\frac{2}{3}\)
b) \(\frac{x}{4}=\frac{-15}{12}=\frac{y+4}{16}\)(có lẽ đề như vậy)
\(\Rightarrow\frac{x}{4}=\frac{-15}{12}\)
\(\Rightarrow12x=-60\)
\(\Rightarrow x=-\frac{60}{12}=-5\)
\(\Rightarrow\frac{x}{4}=\frac{y+4}{16}\)
\(\Rightarrow16x=4y+16\)
\(\Rightarrow-80=4y+16\)
\(\Rightarrow4y=-96\)
\(\Rightarrow y=24\)
câu a hoàn toàn tương tự nên bạn làm nốt nhé!
a)Ta có:
\(\frac{y}{35}=-\frac{6}{14}\Rightarrow y=\frac{-6.35}{14}=-15\)
\(\frac{3}{x}=-\frac{15}{35}\Rightarrow x=\frac{3.35}{-15}=-7\)
Vậy: x=-15;y=-7
b)ta có:
\(\frac{x}{4}=-\frac{15}{12}\Rightarrow x=\frac{-15.4}{12}=-5\)
\(-\frac{15}{12}=\frac{y+4}{16}\Rightarrow y+4=\frac{-15.16}{12}=-20\Rightarrow y=-20-4=-24\)
Vậy x=-5; y=-24
Ta có: \(\frac{-48}{-12}=\frac{12}{x}\Rightarrow x=\frac{\left(-12\right).12}{-48}=3\)
Thế x = 3 \(\Rightarrow\frac{12}{3}=\frac{y^2}{9}\Rightarrow y^2=\frac{12.9}{3}=36\Rightarrow y=\pm6\)
Thế x = 3 \(\Rightarrow\frac{12}{3}=\frac{-256}{t^2}\Rightarrow t^2=\frac{3.\left(-256\right)}{12}=-64\Rightarrow t\in\varnothing\)
Vậy \(x=3;y=\left\{6;-6\right\},t\in\varnothing\)