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c) \(47-\left(x-15\right)=21\)
\(\Leftrightarrow x-15=47-21\)
\(\Leftrightarrow x-15=26\)
\(\Leftrightarrow x=26+15=41\)
d) \(-5-\left(24-x\right)=-11\)
\(\Leftrightarrow24-x=-5+11\)
\(\Leftrightarrow24-x=6\)
\(\Leftrightarrow x=24-6=18\)
Bài 2
a,17-(2+x)=3
(2+x)=17-3
(2+x)=14
x=14-2
x=12
b,Bạn ghi ^ là thế nào
Bài 1:
a)\(21.35-3.25=3.7.7.5-3.5.5=3.5.\left(7.7-5\right)=15.44=660\)
b) \(48.\left(-21\right)+\left(-142\right).\left(-24\right)\)
\(=\left(-2\right).\left(-24\right).\left(-21\right)+\left(-2\right).71.\left(-24\right)\)
\(=\left(-2\right).\left(-24\right).\left(-21+71\right)=48.50=2400\)
c)\(............\)
d)\(=47.\left(45-15-45-15\right)=47.\left(-30\right)=-1410\)
Bài 2:
a)\(17-\left(2+x\right)=3\\ =>2+x=14=>x=12\)
b) ko rõ đề
c)\(4x-15=-75-x\)
\(=>4x+x=-75+15\)
\(=>5x=-60=>x=-12\)
d)\(72-3x=5x+8\)
\(=>-3x-5x=8-72\)
\(=>-8x=-64=>x=8\)
e)\(2\left(x+1\right)+3.\left(x-2\right)=x-4\)
\(=>2x+2+3x-6=x-4\)
\(=>2x+3x-x=-4+6-2\)
\(=>4x=0=>x=0\)
CẬU CS THỂ THAM KHẢO BÀI ALMF TRÊN ĐÂY Ạ. HỌC TỐT NHA:>
Bài 46:
11: Ta có: \(-4\left|x-2\right|=-8\)
\(\Leftrightarrow\left|x-2\right|=2\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=2\\x-2=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=0\end{matrix}\right.\)
Vậy: x∈{0;4}
12: Ta có: \(5\left|x+2\right|=-10\cdot\left(-2\right)\)
\(\Leftrightarrow5\left|x+2\right|=20\)
\(\Leftrightarrow\left|x+2\right|=4\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=4\\x+2=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-6\end{matrix}\right.\)
Vậy: x∈{-6;2}
13: Ta có: \(6\left|x-2\right|=18:\left(-3\right)\)
\(\Leftrightarrow6\left|x-2\right|=-6\)(1)
Ta có: \(\left|x-2\right|\ge0\forall x\)
\(\Rightarrow6\left|x-2\right|\ge0\forall x\)(2)
Ta có: -6<0(3)
Từ (1), (2) và (3) suy ra x∈∅
Vậy: x∈∅
14: Ta có:\(-7\left|x+4\right|=21:\left(-3\right)\)
\(\Leftrightarrow-7\left|x+4\right|=-7\)
\(\Leftrightarrow\left|x+4\right|=1\)
\(\Leftrightarrow\left[{}\begin{matrix}x+4=1\\x+4=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-5\end{matrix}\right.\)
Vậy: x∈{-5;-3}
15: Ta có: \(4\left|x+1\right|=8\left(-2\right)-8\left(-5\right)\)
\(\Leftrightarrow4\left|x+1\right|=-16-\left(-40\right)\)
\(\Leftrightarrow4\left|x+1\right|=24\)
\(\Leftrightarrow\left|x+1\right|=6\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=6\\x+1=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-7\end{matrix}\right.\)
Vậy: x∈{-7;5}
16: Ta có: \(3\left|x+5\right|=-9\)(4)
Ta có: |x+5|≥0∀x
⇒3|x+5|≥0∀x(5)
Ta có: -9<0(6)
Từ (4), (5) và (6) suy ra x∈∅
Vậy: x∈∅
17: Ta có: \(-8\left|x-3\right|=24-16:2\)
\(\Leftrightarrow-8\left|x-3\right|=16\)
\(\Leftrightarrow\left|x-3\right|=-2\)
mà |x-3|≥0>-2∀x
nên x∈∅
Vậy: x∈∅
18: Ta có: \(-3\left|x+6\right|=6\cdot2-9\)
\(\Leftrightarrow-3\left|x+6\right|=3\)
\(\Leftrightarrow\left|x+6\right|=-1\)
mà |x+6|≥0>-1∀x
nên x∈∅
Vậy: x∈∅
19: Ta có: \(5-\left|x+7\right|=4\)
\(\Leftrightarrow\left|x+7\right|=1\)
\(\Leftrightarrow\left[{}\begin{matrix}x+7=-1\\x+7=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-8\\x=-6\end{matrix}\right.\)
Vậy: x∈{-8;-6}
20: Ta có: \(12-\left|x+8\right|=10\)
\(\Leftrightarrow\left|x+8\right|=2\)
\(\Leftrightarrow\left[{}\begin{matrix}x+8=2\\x+8=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-6\\x=-10\end{matrix}\right.\)
Vậy: x∈{-10;-6}
1) 5.( x - 6 ) - 2.( x + 9 ) = 21
5x - 30 - 2x - 18 = 21
3x - 48 = 21
3x = 21 + 48
3x = 69
x = 23
2) 2.( x + 3 ) + 3.( x + 1 ) = 15 - ( - 9 )
2x + 6 + 3x + 3 = 24
5x + 9 = 24
5x = 24 - 9
5x = 15
x = 3
3) ( - x + 5 ).(3 - x ) = 0
=> - x + 5 = 0 hoặc 3 - x = 0
=> x = 5 hoặc x = 3
4) ( x - 12 ) - 15 = ( 20 - 7 ) - ( 18 + x )
x - 12 - 15 = 13 - 18 - x
x - 27 = - 5 - x
x + x = - 5 + 27
2x = 22
x = 11
5) x - ( 17 - 8 ) = 5 + ( 10 - 3x )
x - 9 = 5 + 10 - 3x
x + 3x = 15 + 9
4x = 24
x = 6
a) (x+2)-(x+8)=17.(-2)
-> (x+2)-(x+8)= -34
-> +) x+2 = -34
+) x+8 = -34
-> +) x= (-34)-2
+) x= (-34)-8
-> +) x= -36
+) x= -42
Vậy x E { -36; -42 }
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