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1 . 56- (-23) =56+23 =79
2. 45- 67 = -22
3. -34-59 =-93
4. 81+(-41)=40
( + 56 ) - ( - 23 ) = 79
45 - 67 = -22
- 34 - 59 = - 93
81 + ( - 41 ) = 40
\(27-\left(14-x\right)=-x+3\)
\(\Leftrightarrow\)\(27-14+x=-x+3\)
\(\Leftrightarrow\)\(13+x=-x+3\)
\(\Leftrightarrow\)\(2x=-10\)
\(\Leftrightarrow\)\(x=-5\)
Vậy...
Câu 1 :
=> 27-14+x = -x+3
=> 13+x = -x+3
=> x = -x+3-13 = -x-10
=> 10 = -x-x = -2x
=> x = 10 : (-2) = -5
Vậy x = -5
Tk mk nha
\(4\left(x-2\right)+10=5-3x\)
\(\Rightarrow4x-8+10=5-3x\)
\(\Rightarrow4x+3x=5-10+8\)
\(\Rightarrow7x=3\)
\(\Rightarrow x=\frac{3}{7}\)
\(4x-8-10=7-x\)
\(\Rightarrow4x+x=7+10+8\)
\(\Rightarrow5x=25\)
\(\Rightarrow x=5\)
\(\left|2x-1\right|=3\)
\(\Leftrightarrow\orbr{\begin{cases}2x-1=3\\2x-1=-3\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\\x=-1\end{cases}}}\)
\(\left|1-2x\right|=5\)
\(\Leftrightarrow\orbr{\begin{cases}1-2x=5\\1-2x=-5\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-2\\x=3\end{cases}}}\)
\(\left|4+x\right|=10\)
\(\Leftrightarrow\orbr{\begin{cases}4+x=10\\4+x=-10\end{cases}\Leftrightarrow\orbr{\begin{cases}x=6\\x=-14\end{cases}}}\)
A=\(\frac{\left(2^2\right)^5\cdot\left(3^3\right)^4-2\cdot2^9\cdot3^9}{2^{10}\cdot3^8+2^8\cdot3^8\cdot2^2\cdot5}=\frac{2^{10}\cdot3^{12}-2^{10}\cdot3^9}{2^{10}\cdot3^8+2^{10}\cdot3^8\cdot5}=\frac{2^{10}\cdot3^8\left(3^4-3\right)}{2^{10}\cdot3^8\left(1+5\right)}=\frac{78}{6}=13\)
\(\frac{10}{3}\) . x \(+\)\(\frac{67}{4}\) \(=\)-13,25
\(\frac{10}{3}\) . x \(+\)\(\frac{67}{4}\) \(=\)\(\frac{-53}{4}\)
\(\frac{10}{3}\) . x \(=\)\(\frac{-53}{4}\)\(-\)\(\frac{67}{4}\)
\(\frac{10}{3}\) . x \(=\)\(-\)30
x \(=\)\(-\)30 \(\div\)\(\frac{10}{3}\)
x \(=\)\(-\)9
Ta có:
\(\frac{x}{4}-\frac{1}{y}=\frac{1}{2}\)
\(\frac{xy}{4y}-\frac{4}{4y}=\frac{1}{2}\)
\(\frac{xy-4}{4y}=\frac{1}{2}\)
\(2.\left(xy-4\right)=4y\)
\(2xy-8-4y=0\)
\(2xy-2-4-4y=0\)
\(2.\left(xy+1\right)-4.\left(y+1\right)=0\)
\(2.\left(xy+1\right)-2.2.\left(y+1\right)=0\)
\(2.\left[\left(xy+1\right)-2.\left(y+1\right)\right]=0\)
\(xy+1-2y-2=0\)
\(y.\left(x-2\right)=1\)
Ta có:1=1.1=(-1).(-1)
Do đó ta có bảng sau:
y | 1 | -1 |
x-2 | 1 | -1 |
x | 3 | 1 |
Vậy cặp (x;y) TM là:(3;1)(1;-1)
Theo đầu bài ta có:
\(\frac{x}{4}-\frac{1}{y}=\frac{1}{2}\)
\(\Rightarrow\frac{xy-4}{4y}=\frac{1}{2}\)
\(\Rightarrow2\left(xy-4\right)=4y\)
\(\Rightarrow xy-4=2y\)
\(\Rightarrow xy-2y=4\)
\(\Rightarrow y\left(x-2\right)=4\)
Từ đó ta có bảng sau:
y | -4 | -1 | 1 | 4 |
x - 2 | -1 | -4 | 4 | 1 |
x | 1 | -2 | 6 | 3 |
Trả lời:
10/3 *x+67/4=-13,25
10/3 x= -30
x= -9