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a, (10/3:x).(-5/4)=-10/3
10/3:x=-10/3:(-5/4)
10/3:x=8/3
x=10/3:8/3
x=5/4
b,(-6/5+x):(-18/5)=-1/4
-6/5+x=-1/4.(-18/5)
-6/5+x=9/10
x=9/10-(-6/5)=9/10+6/5
x=21/10
c,-22/15.x+1/3=-2/5
-22/15.x=-2/5-1/3=-11/15
x=-11/15:(-22/15)
x=11/21
d,(0,25-30%x).1/3=-31/6+1/4=-59/12
1/4-3/10x=-59/12:1/3=-59/4
3/10x=1/4-(-59/4)=1/4+59/4=15
x=15:3/10
x=50
e,(0,5x-3/7):1/2=8/7
1/2x=8/7.1/2=4/7
x=4/7:1/2
x=8/7
a: =7/8:(2/9-18+1/36)-5/12
=-7/142-5/12=-397/852
b: =3/7(4/9+5/9:6/12)=2/3
c: =5^8(16/31-47/31)+1/3=-5^8+1/3
d: =7/2(3/8+5/8:4/15)=609/64
\(a,\left(x-\dfrac{5}{8}\right).\dfrac{5}{8}=-\dfrac{15}{36}\)
\(\left(x-\dfrac{5}{8}\right)=-\dfrac{15}{36}\div\dfrac{5}{8}\)
\(x-\dfrac{5}{8}=-\dfrac{2}{3}\)
\(x=-\dfrac{2}{3}+\dfrac{5}{8}\)
\(x=-\dfrac{1}{24}\)
\(b,\left(x-\dfrac{1}{3}\right)=\dfrac{5}{6}\)
\(\Rightarrow x-\dfrac{1}{3}=\dfrac{5}{6}\)
\(x=\dfrac{5}{6}+\dfrac{1}{3}\)
\(x=\dfrac{7}{6}\)
\(a,\left(x-\dfrac{5}{8}\right)\cdot\dfrac{8}{18}=-\dfrac{15}{16}\\ x-\dfrac{5}{8}=-\dfrac{15}{36}:\dfrac{8}{18}\\ x-\dfrac{5}{8}=-\dfrac{15}{16}\\ x=-\dfrac{15}{16}+\dfrac{5}{8}\\ x=-\dfrac{15}{16}+\dfrac{10}{16}\\ x=-\dfrac{5}{16}\\ b,x-\dfrac{1}{3}=\dfrac{5}{6}\\ x=\dfrac{5}{6}+\dfrac{1}{3}\\ x=\dfrac{5}{6}+\dfrac{2}{6}\\ x=\dfrac{7}{6}\)
\(\text{- ( 2789 _ 435 ) + ( 1789 _ 1435 )}\)
\(=-2789+435+1789-1435\)
\(=\left(-2789+1789\right)+\left(435-1435\right)\)
\(=-1000+-1000\)
\(=-2000\)
\(=-\left(-2010\right)+36.41-36.\left(-59\right)\)
\(=2010+36.\left(41+59\right)\)
\(=2010+36.100\)
\(=2010+3600\)
\(=5610\)
\(-75.\left(18-65\right)-65.\left(75-18\right)\)
\(=-75.18+75.65-65.75+65.18\)
\(=18.\left(-75+65\right)+75.\left(65-65\right)\)
\(=18.\left(-10\right)+75.0\)
\(=-180\)
\(-15:x=3\)
\(x=-15:3\)
\(x=-5\)
\(-3x+8=7\)
\(-3x=-1\)
\(x=\frac{1}{3}\)
\(\left(x-6\right).\left(7-x\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-6=0\\7-x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=6\\x=7\end{cases}}}\)
\(\Rightarrow x\in\left\{6;7\right\}\)
\(2.\left(x-3\right)-3.\left(x-5\right)=4.\left(3-x\right)-18\)
\(2x-6-3x+15=12-4x-18\)
\(2x-3x+4x=12-18-15+6\)
\(3x=-15\)
\(\Rightarrow x=-5\)
\(-a.\left(c-d\right)-d.\left(a+c\right)=-c.\left(a+d\right)\)
\(-a.c+a.d-d.a+-d.c=-c.\left(a+d\right)\)
\(-c.\left(a+d\right)+a.\left(d-d\right)=-c.\left(a+d\right)\)
\(-c.\left(a+d\right)+a.0=-c.\left(a+d\right)\)
\(\Rightarrow-c.\left(a+d\right)=-c.\left(a+d\right)\)
(3a+2).(2a–1)+(3–a).(6a+2)–17.(a–1)
=6a²−3a+4a−2+18a+6−6a²−2a−17a+17
=(6a²−6a²)+(−3a+4a+18a−2a−17a)+(17−2+6)
=0+0+21
=21
học tốt
Câu 1:
a) Ta có: \(\left(x-\frac{5}{8}\right)\cdot\frac{5}{18}=\frac{15}{36}\)
\(\Leftrightarrow x-\frac{5}{8}=\frac{15}{36}:\frac{5}{8}=\frac{15}{36}\cdot\frac{8}{5}=\frac{120}{180}=\frac{2}{3}\)
\(\Leftrightarrow x=\frac{2}{3}+\frac{5}{8}=\frac{16}{24}+\frac{15}{24}=\frac{41}{24}\)
Vậy: \(x=\frac{41}{24}\)
b) Ta có: \(\left|x-\frac{1}{3}\right|=\frac{5}{6}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\frac{1}{3}=\frac{5}{6}\\x-\frac{1}{3}=-\frac{5}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{5}{6}+\frac{1}{3}=\frac{5}{6}+\frac{2}{6}=\frac{7}{6}\\x=\frac{-5}{6}+\frac{1}{3}=\frac{-5}{6}+\frac{2}{6}=\frac{-3}{6}=\frac{-1}{2}\end{matrix}\right.\)
Vậy: \(x\in\left\{\frac{7}{6};\frac{-1}{2}\right\}\)