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\(\frac{a}{27}=\frac{-5}{9}=\frac{-45}{b}\)
\(=>\frac{a.3}{81}=\frac{-45}{81}=\frac{-45}{b}\)
=>b=81
a.3=-45
a=-15
Vậy a=-15;b=81
x(x+2)=0
suy ra x=0 hoặc x+2=0
5-2x=-7
2x=-7+5
2x=-(7-5)
2x=-2
x=-2:2
x=-1
Vậy x=-1
NHỚ TÍCH MK NHA
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Jim Rohn – Triết lý cuộc đời
\(a,2.\left(x-5\right)-3.\left(x+7\right)=14\)
\(2x-10-3x-21=14\)
\(-x-31=14\)
\(x=-31-14\)
\(x=-45\)
\(b,5.\left(x-6\right)-2\left(x+3\right)=12\)
\(5x-30-2x-6=12\)
\(3x-36=12\)
\(3x=12+36\)
\(3x=48\)
\(x=16\)
\(c,-5.\left(2-x\right)+4.\left(x-3\right)=10.x-15\)
\(-10+5x+4x-12=10x-15\)
\(-6x-22=10x-15\)
\(-6x-10x=-15+22\)
\(-16x=7\)
\(x=-\frac{7}{16}\)
Câu d , e f tương tự nha
nhiều quá :((
\(a,2\left(x-5\right)-3\left(x+7\right)=14\)
\(2x-10-3x-21=14\)
\(-x-31=14\)
\(-x=45\)
\(x=45\)
\(b,5\left(x-6\right)-2\left(x+3\right)=12\)
\(5x-30-2x-6=12\)
\(3x-36==12\)
\(3x=48\)
\(x=16\)
\(c,3\left(x-4\right)-\left(8-x\right)=12\)
\(3x-12-8+x=0\)
\(4x-20=0\)
\(4x=20\)
\(x=5\)
Cố nốt nha bn !
cảm ơn, bn nha:)))
mà hình như bạn TOP 3 trả lời câu hỏi pải ko nhỉ???
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
-37 + 14 +26 + 37
=-37 + 37 + 14 + 36
=0 + 50
=50
-24 + 6 + 10 + 24
=(-24 + 24) +(6+10)
= 0+16
= 16
b)
\(\dfrac{1}{2}\left(\dfrac{4}{5}-\dfrac{3}{2}\right)+x=5\left(x-\dfrac{1}{3}\right)\)
=> \(-\dfrac{7}{20}+x=5x-\dfrac{5}{3}\)
=> \(\dfrac{79}{60}=4x\)
=> \(\dfrac{79}{240}=x\)
Vậy \(\dfrac{79}{240}=x\)
c)
\(2\left(x-5\right)-3\left(x+7\right)=14\)
=> \(2x-10-3x-21=14\)
=> \(-x-31=14\)
=> \(-x=45\)
=> x = -45
Vậy x là -45
d)
-8 . |x - 3| = -32
=> |x - 3| = 4
=> \(\left\{{}\begin{matrix}x-3=4\\x-3=-4\end{matrix}\right.\)
=> x = 7 và x = -1
Vậy x \(\in\left\{-1;7\right\}\)
e)
-90 : |x + \(\dfrac{3}{4}\)| = -45
=> |x + \(\dfrac{3}{4}\)| = -90 : -45 = 2
=> \(\left\{{}\begin{matrix}x+\dfrac{3}{4}=2\\x+\dfrac{3}{4}=-2\end{matrix}\right.\)
=> x = \(\dfrac{5}{4}\) và x = \(-\dfrac{11}{4}\)
Vậy x \(\in\left\{-\dfrac{11}{4};\dfrac{5}{4}\right\}\)
Giải:
b, \(\dfrac{1}{2}.\left(\dfrac{4}{5}-\dfrac{3}{2}\right)+x=5.\left(x-\dfrac{1}{3}\right)\)
\(\Rightarrow\dfrac{1}{2}.\left(\dfrac{8}{10}-\dfrac{8}{10}\right)+x=5x-\dfrac{5}{3}\)
\(\Rightarrow\dfrac{1}{2}.0+x=5x-\dfrac{5}{3}\)
\(\Rightarrow x=5x-\dfrac{5}{3}\)
\(\Leftrightarrow5x-\dfrac{5}{3}=x\)
\(\Rightarrow5x-x=\dfrac{5}{3}\)
\(\Rightarrow4x=\dfrac{5}{3}\)
\(\Rightarrow x=\dfrac{5}{12}\)
c, 2.(x - 5) - 3.(x + 7) = 14
\(\Rightarrow\left(2x-10\right)-\left(3x+21\right)=14\)
\(\Rightarrow\) 2x - 10 - 3x - 21 = 14
\(\Rightarrow-x-31=14\)
\(\Rightarrow-x=45\)
\(\Rightarrow x=-45\)
d, \(-8.\left|x-3\right|=-32\)
\(\Rightarrow\left|x-3\right|=4\)
\(\Rightarrow\left[{}\begin{matrix}x-3=4\\x-3=-4\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=7\\x=-1\end{matrix}\right.\)
e, \(-90:\left|x+\dfrac{3}{4}\right|=-45\)
\(\Rightarrow\left|x+\dfrac{3}{4}\right|=2\)
\(\Rightarrow\left[{}\begin{matrix}x+\dfrac{3}{4}=2\\x+\dfrac{3}{4}=-2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{11}{4}\\x=-\dfrac{1}{4}\end{matrix}\right.\)