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a) -3x+4+5x=-10-x
-3x+4+5x+10+x=0
(-3x+5x+x)+10=0
3x+10=0
3x=-10
x=\(\dfrac{-10}{3}\)
Vậy x=\(\dfrac{-10}{3}\)
b)-x+1=-3x-8
-x+1+3x+8=0
(-x+3x)+(1+8)=0
2x+9=0
2x=-9
x=\(\dfrac{-9}{2}\)
Vậy x=\(\dfrac{-9}{2}\)
c)8-(x-1)=10+(x+5)
8-x+1=10+x+5
9-x=15+x
9-x-15-x=0
(9-15)-(x+x)=0
-6-2x=0
2x=-6
x=-3
Vậy x=-3
d)100+(x+7)-(-2x+3)=8+(x+100)
100+x+7+2x-3=8+x+100
(x+2x)+(100+7-3)=(8+100)+x
3x+104=108+x
3x+104-108-x=0
(3x-x)+(104-108)=0
2x-4=0
2x=4
x=2
Vậy x=2
e, \(\left|2x+5\right|=\left|x-1\right|\)
\(\Rightarrow\left\{{}\begin{matrix}2x+5=1-x\\2x+5=x-1\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}3x=-4\\x=-6\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{4}{3}\\x=-6\end{matrix}\right.\)
g, \(\left|-x+4\right|=\left|-3x-8\right|\)
\(\Rightarrow\left\{{}\begin{matrix}-x+4=3x+8\\-x+4=-3x-8\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}-4x=4\\2x=-12\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-1\\x=-6\end{matrix}\right.\)
h, \(\left|x+4\right|=\left|-3-8\right|\)
\(\Rightarrow\left|x+4\right|=\left|-11\right|=11\)
\(\Rightarrow\left\{{}\begin{matrix}x+4=-11\\x+4=11\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-15\\x=7\end{matrix}\right.\)
Chúc bạn học tốt!!!
a, x + \(\dfrac{1}{5}\) = \(\dfrac{5}{6}\)
x = \(\dfrac{5}{6}\) - \(\dfrac{1}{5}\)
x = \(\dfrac{19}{30}\)
b, x - \(\dfrac{3}{5}=\dfrac{6}{7}\)
x = \(\dfrac{6}{7}+\dfrac{3}{5}\)
x = \(1\dfrac{16}{35}\)
c, - x - \(\dfrac{7}{5}=\dfrac{-8}{9}\)
- x = \(\dfrac{-8}{9}+\dfrac{7}{5}\)
x = \(\dfrac{23}{45}\)
d, \(\dfrac{3}{8}-x=\dfrac{2}{3}\)
\(x=\dfrac{3}{8}-\dfrac{2}{3}\)
\(x=\dfrac{-7}{24}\)
e, \(|x-\dfrac{8}{9}|=\dfrac{1}{3}\)
\(x=\dfrac{1}{3}+\dfrac{8}{9}\)
\(x=\dfrac{-11}{9}\)
Hiệu giữa SBT mới và cũ là:
353 – 23 = 330
Hiệu số phần bằng nhau là:
3-1 = 2 phần
Số bị trừ cũ là: 330 : 2 = 165
Số trừ cũ là : 165- 23 = 142
a: -3x+4+5x=-10-x
=>2x+4=-x-10
=>3x=-14
hay x=-14/3
b: \(-x+1=-3x-8\)
=>-x+3x=-8-1
=>2x=-9
hay x=-9/2
c: \(8-\left(x-1\right)=10+\left(x+5\right)\)
=>x+15=8-x+1
=>x+15=9-x
=>2x=-6
hay x=-3
d: \(100+\left(x+7\right)-\left(-x+3\right)=8+\left(x+100\right)\)
=>x+7+x-3=8+x
=>2x+4-x-8=0
=>x=4
a) \(x^3=-8\)
\(\Rightarrow x=-2\)
b) \(x^{10}-x^8=0\)
\(\Leftrightarrow x^{10}=x^8\)
\(\Rightarrow x=1\)
\(\Rightarrow x=-1\)
\(\Rightarrow x=0\)
c) \(\left(x+1\right)^5-\left(x+1\right)^2=0\)
\(\Leftrightarrow\left(x+1\right)^5=\left(x+1\right)^2\)
\(thay\) x vào thì ta sẽ có giá trị thỏa mãn.
d) \(2^5-4=2^n-1\)
\(\Rightarrow32-4=2^{n-1}\)
\(\Rightarrow28=2^{n-1}\)
\(\leftrightarrow\) \(n\) không có giá trị thỏa mãn
a) x3 = -8
=> x = -2
b) x10 - x8 = 0
x10 = x8
=> x = 1 ; x = -1 ; x = 0
a) -3(x-4)+5(x-1)=-7
=>-3x+12+5x-5=-7
=>2x+7=-7
=>2x=-14=>x=-7
b) -4./x-8/+12=0
=>/x-8/=3
=>x-8=3 hoặc -3
(tự tính)
\(a,x-7\frac{5}{8}=1\frac{1}{4}\)
=> \(x-\frac{61}{8}=\frac{5}{4}\)
=> \(x=\frac{5}{4}+\frac{61}{8}\)
=> \(x=\frac{10}{8}+\frac{61}{8}=\frac{71}{8}=8\frac{7}{8}\)
\(b,x+7\frac{5}{8}=9\frac{1}{4}\)
=> \(x+\frac{43}{5}=\frac{37}{4}\)
=> \(x=\frac{37}{4}-\frac{43}{5}=\frac{13}{20}\)
\(c,\left[x-7\frac{5}{8}\right]:\frac{1}{2}=3\)
=> \(\left[x-\frac{61}{8}\right]=3\cdot\frac{1}{2}\)
=> \(\left[x-\frac{61}{8}\right]=\frac{3}{2}\)
=> \(x-\frac{61}{8}=\frac{3}{2}\)
=> \(x=\frac{3}{2}+\frac{61}{8}=\frac{12}{8}+\frac{61}{8}=\frac{73}{8}=9\frac{1}{8}\)
d, \(\frac{x}{1\cdot3}+\frac{x}{3\cdot5}+\frac{x}{5\cdot7}+...+\frac{x}{97\cdot99}=99\)
=> \(\frac{x}{2}\left[\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{97\cdot99}\right]=99\)
=> \(\frac{x}{2}\left[1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{97}-\frac{1}{99}\right]=99\)
=> \(\frac{x}{2}\left[1-\frac{1}{99}\right]=99\)
=> \(\frac{x}{2}\cdot\frac{98}{99}=99\)
=> \(\frac{98x}{198}=99\)
=> 98x = 99 . 198
=> 98x = 19602
=> x = 19602 : 98 = 9801/49
a) \(x-7\frac{5}{8}=1\frac{1}{4}\)
=> \(x=\frac{5}{4}+\frac{61}{8}\)
=> \(x=\frac{71}{8}\)
b) \(x+7\frac{5}{8}=9\frac{1}{4}\)
=> \(x=\frac{37}{4}-\frac{61}{8}\)
=> \(x=\frac{13}{8}\)
c) \(\left(x-7\frac{5}{8}\right):\frac{1}{2}=3\)
=> \(x-\frac{61}{8}=3.\frac{1}{2}\)
=> \(x-\frac{61}{8}=\frac{3}{2}\)
=> \(x=\frac{3}{2}+\frac{61}{8}\)
=> \(x=\frac{73}{8}\)
d) \(\frac{x}{1.3}+\frac{x}{3.5}+...+\frac{x}{97.99}=99\)
=> \(x.\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{97.99}\right)=99\)
=> \(\frac{1}{2}x\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{97}-\frac{1}{99}\right)=99\)
=> \(x\left(1-\frac{1}{99}\right)=99:\frac{1}{2}\)
=> \(x.\frac{98}{99}=198\)
=> \(x=198:\frac{98}{99}=\frac{9801}{49}\)
a)\(x-\frac{1}{4}=\frac{5}{8}\)
\(x=\frac{5}{8}+\frac{1}{4}\)
\(x=\frac{7}{8}\)
b)\(\frac{x}{12}=\frac{-1}{24}-\frac{1}{8}\)
\(\frac{x}{12}=\frac{-1}{24}+\frac{-1}{8}\)
\(\frac{x}{12}=\frac{-1}{6}\)
\(x=\frac{-1}{6}\times12\)
\(x=-2\)
a)x-1/4=5/8
=>x=5/8+1/4
x=7/8
b) x/12=-1/24-1/8
=>x/12=-1/6
=>x=-2
c).......... Tự làm
a)15/8+3/4-5/12
=45+18-10/24
=53/24
b)11/24.12/33+5/6
=11.12/12.2.11.3+5/6
=1/6+5/6
=6/6=1
c)15/8+7/24:5/8
=15/8+7/24.8/5
=15/8+7.8/3.8.5
=15/8+7/15
=đề sai, nếu đúng thì như này
=8/15+7/15
=15/15=1
ĐKXĐ: x<>1
\(\dfrac{x-1}{8}=\dfrac{8}{x-1}\)
=>\(\left(x-1\right)\cdot\left(x-1\right)=8\cdot8\)
=>\(\left(x-1\right)^2=64\)
=>\(\left[{}\begin{matrix}x-1=8\\x-1=-8\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=8+1=9\left(nhận\right)\\x=-8+1=-7\left(nhận\right)\end{matrix}\right.\)
\(\dfrac{x-1}{8}\) = \(\dfrac{8}{x-1}\) (đk \(x-1\ne0\) ⇒ \(x\ne\) 1)
(\(x-1\)).(\(x-1\)) = 8.8
(\(x-1\))2 = 82
\(\left[{}\begin{matrix}x-1=-8\\x-1=8\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=-8+1\\x=8+1\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=-7\\x=9\end{matrix}\right.\)
Vậy \(x\) \(\in\) {-7; 9)