Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(a)484+x=\left(-632\right)+\left(-548\right)\)
\(484+x=-1180\)
\(x=(-1180)-484\)
\(x=-1664\)
\(b)\left(-x\right)+\left(-62\right)+\left(-46\right)=-14\)
\(\left(-x\right)+\left(-62\right)=-14-\left(-46\right)\)
\(\left(-x\right)+\left(-62\right)=-14+46\)
\(\left(-x\right)+\left(-62\right)=32\)
\(\left(-x\right)=32-\left(-62\right)\)
\(\left(-x\right)=32+62\)
\(\left(-x\right)=94\)
\(\Rightarrow x=94\)
\(c)25+\left(x-5\right)=-415-\left(15-415\right)\)
\(25+\left(x-5\right)=-415-\left(-400\right)\)
\(25+\left(x-5\right)=-415+400\)
\(25+\left(x-5\right)=-15\)
\(x-5=-15-25\)
\(x-5=-40\)
\(x=-40+5\)
\(x=-35\)
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???
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}
\(a,2\left(x-5\right)-3\left(x+7\right)=14\)
\(\Leftrightarrow2x-10-3x-21=14\)
\(\Leftrightarrow-x-31=14\Leftrightarrow-x=45\Leftrightarrow x=-45\)
\(c,5\left(x-6\right)-2\left(x+3\right)=12\)
\(\Leftrightarrow5x-30-2x-6=12\)
\(\Leftrightarrow3x-36=12\Leftrightarrow3x=48\Leftrightarrow x=16\)
Các câu còn lại tương tự
a, 2(x - 5) - 3(x + 7) = 14\(\Rightarrow2\chi-10-3\chi+21=14\Rightarrow2\chi-3\chi=14+10-21\)
\(\Rightarrow-\chi=3\Rightarrow\chi=3\)
Bài 1:
1; ( - 35) : (-7) = 5
2; (- 42) : 21 = - 2
3; 45 : (-9) = -5
4; 18 : 9 = 2
5; (- 30) : (- 15) = 2
6; 0 : 18 = 0
7; 0 : (-13) = 0
8; 44 : (-4) = - 11
9; - 55 : 11 = - 5
10; 46 : 23 = 2
\(-5.\left(2-x\right)+4\left(x-3\right)=10x+15\)
\(-10+5x+4x-12=10x+15\)
\(9x-22=10x+15\)
\(10x-9x=-22-15\)
\(x=-37\)
\(7.\left(x-9\right)-5\left(6-x\right)=-6+11x\)
\(7x-63-30+5x=-6+11x\)
\(12x-93=-6+11x\)
\(12x-11x=-6+93\)
\(x=87\)
\(xy+14+2y+7x=-10\)
\(\left(xy+2y\right)+\left(14+7x\right)=-10\)
\(y\left(x+2\right)+7\left(2+x\right)=-10\)
\(\left(x+2\right)\left(y+7\right)=-10\)
Câu cuối mk chỉ biết làm đến đó thôi
bn tự làm nha
a, \(\dfrac{62}{7}.x=\dfrac{29}{90}.\dfrac{3}{56}\)
\(\dfrac{62}{7}.x=\dfrac{29}{1680}\)
\(x=\dfrac{29}{1680}:\dfrac{62}{7}\)
\(x=\dfrac{29}{14880}\)
b, \(\dfrac{1}{5}:x=\dfrac{1}{5}-\dfrac{1}{7}\)
\(\dfrac{1}{5}:x=\dfrac{2}{35}\)
\(x=\dfrac{1}{5}:\dfrac{2}{35}\)
\(x=\dfrac{7}{2}\)
c, \(\left(x+\dfrac{1}{4}-\dfrac{1}{3}\right):\left(2+\dfrac{1}{6}-\dfrac{1}{4}\right)=\dfrac{7}{46}\)
\(\left(x+\dfrac{1}{4}-\dfrac{1}{3}\right):\left(\dfrac{13}{6}-\dfrac{1}{4}\right)=\dfrac{7}{46}\)
\(\left(x+\dfrac{1}{4}-\dfrac{1}{3}\right):\dfrac{23}{12}=\dfrac{7}{46}\)
\(\left(x+\dfrac{-1}{12}\right):\dfrac{23}{12}=\dfrac{7}{46}\)
\(\left(x+\dfrac{-1}{12}\right)=\dfrac{7}{46}.\dfrac{23}{12}\)
\(x+\dfrac{-1}{12}=\dfrac{7}{24}\)
\(x=\dfrac{7}{24}-\dfrac{-1}{12}\)
\(x=\dfrac{3}{8}\)
- 62 + x = ( - 14 ) + 46 - x
x + x = ( - 14 ) + 46 + 62
2x = 94
x = 47
\(-62+x=\left(-14\right)+46-x\\ \Rightarrow x-62=32-x\\ \Rightarrow x-62-32+x=0\\ \Rightarrow2x-94=0\\ \Rightarrow2x=94\\ \Rightarrow x=47\\ b,5^{x+2}-5^x=10^2.6\\ \Rightarrow25.5^x-5^x=100.6\\ \Rightarrow24.5^x=600\\ \Rightarrow5^x=25\\ \Rightarrow5^x=5^2\\ \Rightarrow x=2\)