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\(\frac{27}{4}=\frac{-x}{3}=>x=-\frac{81}{4}\notinℤ\)
\(^{y^2=\frac{4}{9}=\left(\frac{2}{3}\right)^2=>y=\pm\frac{2}{3}\notinℤ}\)
\(\frac{27}{4}=\frac{\left(z+3\right)}{-4}=\left(z+3\right)=-27=\left(-3\right)^3=>z+3=-3=>z=-6\)
\(+)|t|-2=-54=>|t|=-52\)(vô lí)
\(+)|t|-2=54=>|t|=56=>t=\pm56\)
Bài 1:
a: \(\Leftrightarrow\left|x+\dfrac{4}{15}\right|=-2.15+3.75=\dfrac{8}{5}\)
=>x+4/15=8/5 hoặc x+4/15=-8/5
=>x=4/3 hoặc x=-28/15
b: \(\Leftrightarrow\left[{}\begin{matrix}\dfrac{5}{3}x=-\dfrac{1}{6}\\\dfrac{5}{3}x=\dfrac{1}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-1}{6}:\dfrac{5}{3}=\dfrac{-3}{30}=\dfrac{-1}{10}\\x=\dfrac{1}{10}\end{matrix}\right.\)
c: \(\Leftrightarrow\left|x-1\right|-1=1\)
=>|x-1|=2
=>x-1=2 hoặc x-1=-2
=>x=3 hoặc x=-1
Bài 2:
b: \(\Leftrightarrow\left\{{}\begin{matrix}x-y=0\\y+\dfrac{9}{25}=0\end{matrix}\right.\Leftrightarrow x=y=-\dfrac{9}{25}\)
Bài 3:
a: \(A=\left|x+\dfrac{15}{19}\right|-1>=-1\)
Dấu '=' xảy ra khi x=-15/19
b: \(\left|x-\dfrac{4}{7}\right|+\dfrac{1}{2}>=\dfrac{1}{2}\)
Dấu '=' xảy ra khi x=4/7
a. \(\dfrac{1}{2}-\left(\dfrac{1}{3}+\dfrac{3}{4}\right)\le x\le\dfrac{1}{24}.\left(\dfrac{1}{3}-\dfrac{1}{3}\right)\)
\(\dfrac{1}{2}-\dfrac{13}{12}\le x\le\dfrac{1}{24}.0\) ( lười viết nên điền kết quả luôn )
\(\dfrac{-7}{12}\le x\le0\)
\(0,5833...\le x\le0\)
Vì \(x\in Z\)\(\Rightarrow x\in\left\{0\right\}\)
Vậy...
b. \(-4\dfrac{1}{3}\left(\dfrac{1}{2}+\dfrac{1}{6}\right)\le x\le\dfrac{-2}{3}\left(\dfrac{1}{3}-\dfrac{1}{2}.\dfrac{3}{4}\right)\)
\(\dfrac{-26}{9}\le x\le\dfrac{1}{36}\)
\(-2,8888...\le x\le0,277...\)
Vì \(x\in Z\Rightarrow x\in\left\{-2;-1;0\right\}\)
Vậy ...
a) Để phân số \(\dfrac{3}{n-2}\) là số nguyên thì n - 2 \(⋮\) 3
\(\Rightarrow\) n - 2 \(\in\) Ư(3)
\(\Rightarrow\) n - 2 \(\in\){3; -3; 1;-1}
n \(\in\){5; -1; 3; 2}
c) \(\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+......+\dfrac{1}{28.29}\)
\(=\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+.....+\dfrac{1}{29}-\dfrac{1}{30}\)
\(=\dfrac{1}{3}-\dfrac{1}{30}\)
\(=\dfrac{10}{30}-\dfrac{1}{30}\)
\(=\dfrac{9}{30}\)
=\(\dfrac{3}{10}\)
a: =>-3/2+x-7=5-1/3x+4/15
=>4/3x=413/30
hay x=413/40
b: \(\Leftrightarrow5-\dfrac{3}{2}x=-\dfrac{22}{3}\cdot\dfrac{-11}{8}=\dfrac{121}{12}\)
=>3/2x=-61/12
hay x=-61/18
c: (3x+2)2+|3x+2y|=0
=>3x+2=0 và 3x=-2y
=>x=-2/3 và -2y=-2
=>(x,y)=(-2/3;1)
1) \(-\dfrac{5}{9}+1\dfrac{5}{9}\cdot\left(\dfrac{3}{4}-\dfrac{2}{5}\right):7^2\\ =-\dfrac{5}{9}+\dfrac{14}{9}\cdot\dfrac{7}{20}\cdot\dfrac{1}{49}\\ =-\dfrac{5}{9}+\dfrac{1}{90}=\dfrac{-49}{90}\)
2) \(1\dfrac{13}{15}\cdot0,75-\left(\dfrac{104}{195}+25\%\right)\cdot\dfrac{24}{47}-3\dfrac{12}{13}:3\\ =\dfrac{28}{15}\cdot\dfrac{3}{4}-\left(\dfrac{8}{15}+\dfrac{1}{4}\right)\cdot\dfrac{24}{47}-\dfrac{51}{13}\cdot\dfrac{1}{3}\\ =\dfrac{7}{5}-\dfrac{47}{60}\cdot\dfrac{24}{47}-\dfrac{17}{13}\\ =\dfrac{7}{5}-\dfrac{2}{5}-\dfrac{17}{13}\\ =-\dfrac{4}{13}\)
3) \(1\dfrac{13}{15}\cdot\left(0,5\right)^2\cdot3+\left(\dfrac{8}{15}-1\dfrac{19}{60}\right):1\dfrac{23}{24}\\ =\dfrac{28}{15}\cdot\dfrac{1}{4}\cdot3+\left(\dfrac{8}{15}-\dfrac{79}{60}\right):\dfrac{47}{24}\\ =\dfrac{7}{5}-\dfrac{47}{60}\cdot\dfrac{24}{47}\\ =\dfrac{7}{5}-\dfrac{2}{5}\\ =1\)
Tìm x : 1) \(60\%x+0,4x+x:3=2\\ \Leftrightarrow\dfrac{3}{5}x+\dfrac{2}{5}x+\dfrac{1}{3}x=2\\ \Leftrightarrow\dfrac{4}{3}x=2\\ \Leftrightarrow x=\dfrac{3}{2}\)
Nốt nè bn
\(-2x-\dfrac{-3}{5}:\left(0,5\right)^2=-1\dfrac{1}{4}\\ \Leftrightarrow-2x+\dfrac{3}{5}:\dfrac{1}{4}=-\dfrac{5}{4}\\ \Leftrightarrow-2x+\dfrac{12}{5}=-\dfrac{5}{4}\\ \Leftrightarrow-2x=-\dfrac{73}{20}\\ x=-\dfrac{73}{40}\)
\(\left(\dfrac{2}{3}-x\right):\dfrac{3}{4}=\dfrac{1}{5}\\ \Leftrightarrow\dfrac{2}{3}-x=\dfrac{1}{5}\cdot\dfrac{3}{4}\\ \Leftrightarrow\dfrac{2}{3}-x=\dfrac{3}{20}\\ \Leftrightarrow x=\dfrac{2}{3}-\dfrac{3}{20}\\ \Leftrightarrow x=\dfrac{31}{60}\)
Mấy bài này bạn tự làm đi, chuyển vế tìm x gần giống cấp I mà.
b)\(\dfrac{-3}{5}.x=\dfrac{1}{4}+0,75\)
=>\(\dfrac{-3}{5}.x=1\)
=>\(x=1:\dfrac{-3}{5}\)
=>\(x=\dfrac{-5}{3}\)
Vậy \(x=\dfrac{-5}{3}\)
-x/3=24/4=6
=>x=-18
3/y2=6
=>y2=1/2
hay \(y=\pm\dfrac{\sqrt{2}}{2}\)
\(\dfrac{\left(z+3\right)^3}{-4}=6\)
=>(z+3)3=-24
\(\Leftrightarrow z+3=-\sqrt[3]{24}\)
hay \(z=-\sqrt[3]{24}-3\)
||t|-2|/8=6
=>||t|-2|=48
=>|t|-2=48
=>t=50 hoặc t=-50