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a)Vì \(\hept{\begin{cases}\left|x+19\right|\ge0;\forall x,y\\\left|y-5\right|\ge0;\forall x,y\end{cases}\Rightarrow\left|x+19\right|+\left|y-5\right|\ge0;\forall x,y}\)
\(\Rightarrow\left|x+19\right|+\left|y-5\right|+1890\ge1890;\forall x,y\)
Dấu"="xảy ra \(\Leftrightarrow\hept{\begin{cases}\left|x+19\right|=0\\\left|y-5\right|=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=-19\\y=5\end{cases}}}\)
Vậy Min A=1890 \(\Leftrightarrow\hept{\begin{cases}x=-19\\y=5\end{cases}}\)
b)Vì \(\hept{\begin{cases}-\left|x-7\right|\le0;\forall x,y\\-\left|y+13\right|\le0;\forall x,y\end{cases}}\)\(\Rightarrow-\left|x-7\right|-\left|y+13\right|\le0;\forall x,y\)
\(\Rightarrow-\left|x-7\right|-\left|y+13\right|+1945\le1945;\forall x,y\)
Dấu"="Xảy ra \(\Leftrightarrow\hept{\begin{cases}\left|x-7\right|=0\\\left|y+13\right|=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=7\\y=-13\end{cases}}\)
Vậy Max \(B=1945\Leftrightarrow\hept{\begin{cases}x=7\\y=-13\end{cases}}\)
a) Ta có: \(22x-y=21x+x-y\)
Ta có: \(21x⋮7\)
\(x-y⋮7\)
Do đó: \(21x+x-y⋮7\)(dấu hiệu chia hết của một tổng)
hay \(22x-y⋮7\)(đpcm)
b) Ta có: \(8x+20y=7x+21y+x-y\)
Ta có: \(7x+21y⋮7\)
\(x-y⋮7\)
Do đó: \(7x+21y+x-y⋮7\)(dấu hiệu chia hết của một tổng)
hay \(8x+20y⋮7\)(đpcm)
c) Ta có: \(11x+10y=21x-10x+10y\)
\(=21x-\left(10x-10y\right)\)
Ta có: \(21x⋮7\)
\(10x-10y=10\left(x-y\right)⋮7\)
Do đó: \(21x-\left(10x-10y\right)⋮7\)(dấu hiệu chia hết của một hiệu)
hay \(11x+10y⋮7\)(đpcm)
a) Ta có: 22x - y = 21x + (x - y)
Vì 21 ⋮ 7
Nên 21x ⋮ 7
Mà x - y ⋮ 7
Do đó biểu thức trên ⋮ 7
b) Ta có 8x + 20 = 8x - 8y + 28y = 8(x - y) + 28y
Xét thấy 28 ⋮ 7
Nên 28y ⋮ 7
Và 8(x - y) ⋮ 7
Nên biểu thức trên chia hết cho 7
c)Ta có 11x + 10y = 11x - 11y + 21y = 11(x - y) + 21y
Xét thấy 21 ⋮ 7
Nên 21y ⋮ 7
Và 11(x - y) ⋮ 7
Vậy biểu thức trên chia hết cho 7
1. Tìm \(x\):
a) \(\dfrac{x}{5}=\dfrac{5}{6}+\dfrac{-19}{30}\)
\(\dfrac{x}{5}=\dfrac{1}{5}\)
\(\Rightarrow x=1\)
b) \(\dfrac{-5}{6}-x=\dfrac{7}{12}-\dfrac{1}{3}.x\)
\(\dfrac{-5}{6}-\dfrac{7}{12}=x-\dfrac{1}{3}.x\)
\(x-\dfrac{1}{3}.x=\dfrac{-17}{12}\)
\(\dfrac{2}{3}.x=\dfrac{-17}{12}\)
\(x=\dfrac{-17}{12}:\dfrac{2}{3}\)
\(x=\dfrac{-17}{8}\)
c) \(2016^3.2016^x=2016^8\)
\(2016^x=2016^8:2016^3\)
\(2016^x=2016^{8-3}\)
\(2016^x=2016^5\)
\(\Rightarrow x=5\)
d) \(\left(x+\dfrac{3}{4}\right):\dfrac{5}{2}=3\dfrac{1}{2}\)
\(\left(x+\dfrac{3}{4}\right):\dfrac{5}{2}=\dfrac{7}{2}\)
\(\left(x+\dfrac{3}{4}\right)=\dfrac{7}{2}.\dfrac{5}{2}\)
\(x+\dfrac{3}{4}=\dfrac{35}{4}\)
\(x=\dfrac{35}{4}-\dfrac{3}{4}\)
\(x=\dfrac{32}{4}=8\)
e) \(\left(2,8.x-2^5\right):\dfrac{2}{3}=3^2\)
\(\left(2,8.x-2^5\right)=9.\dfrac{2}{3}\)
\(2,8.x-2^5=6\)
\(2,8.x=6+32\)
\(2,8.x=38\)
\(x=38:2,8\)
\(x=\dfrac{95}{7}\)
f) \(\dfrac{4}{7}.x-\dfrac{2}{3}=\dfrac{2}{5}\)
\(\dfrac{4}{7}.x=\dfrac{2}{5}+\dfrac{2}{3}\)
\(\dfrac{4}{7}.x=\dfrac{16}{15}\)
\(x=\dfrac{16}{15}:\dfrac{4}{7}\)
\(x=\dfrac{28}{15}\)
g) \(\left(\dfrac{3x}{7}+1\right):\left(-4\right)=\dfrac{-1}{28}\)
\(\left(\dfrac{3x}{7}+1\right)=\dfrac{-1}{28}.\left(-4\right)\)
\(\dfrac{3x}{7}+1=\dfrac{1}{7}\)
\(\dfrac{3x}{7}=\dfrac{1}{7}-1\)
\(\dfrac{3x}{7}=\dfrac{-6}{7}\)
\(\Rightarrow3x=-6\)
\(x=\left(-6\right):3\)
\(x=-2\)
2. Thực hiện phép tính:
a) \(\dfrac{1}{2}+\dfrac{1}{2}.\dfrac{2}{3}-\dfrac{1}{3}:\dfrac{3}{4}+1\dfrac{4}{5}\)
\(=\dfrac{1}{2}.\left(\dfrac{2}{3}+1\right)-\dfrac{1}{3}:\dfrac{3}{4}+\dfrac{9}{5}\)
\(=\dfrac{1}{2}.\dfrac{5}{3}-\dfrac{1}{3}:\dfrac{3}{4}+\dfrac{9}{5}\)
\(=\dfrac{5}{6}-\dfrac{4}{9}+\dfrac{9}{5}\)
\(=\dfrac{7}{18}+\dfrac{9}{5}\)
\(=\dfrac{197}{90}\)
b) \(\dfrac{7.5^2-7^2}{7.24+21}\)
\(=\dfrac{7.25-7.7}{7.24+7.3}\)
\(=\dfrac{7.\left(25-7\right)}{7.\left(24+3\right)}\)
\(=\dfrac{7.18}{7.27}\)
\(=\dfrac{2}{3}\)
c) \(\dfrac{2}{3}+\dfrac{1}{3}.\left(\dfrac{-4}{9}+\dfrac{5}{6}\right):\dfrac{7}{12}\)
\(=\dfrac{2}{3}+\dfrac{1}{3}.\dfrac{7}{18}:\dfrac{7}{12}\)
\(=\dfrac{2}{3}+\dfrac{7}{54}:\dfrac{7}{12}\)
\(=\dfrac{2}{3}+\dfrac{2}{9}\)
\(=\dfrac{8}{9}\)
Tim x biet
a) x - \(\frac{5}{7}\)=\(\frac{1}{9}\)
b)\(\frac{-3}{7}\)- x = \(\frac{4}{5}+\frac{-2}{3}\)
\(x-\frac{5}{7}=\frac{1}{9}\)
\(x=\frac{52}{63}\)
\(\frac{-3}{7}-x=\frac{4}{5}+\frac{-2}{3}\)
\(\frac{-3}{7}-x=\frac{2}{15}\)
\(x=\frac{-59}{105}\)
\(=\frac{2a+8-a-7}{5}\)
\(=\frac{a-1}{5}\)
\(=\frac{a+4-5}{5}\)
\(=\frac{a+4}{5}-1\)
mà \(1\in Z\) nên \(\frac{a+4}{5}\in Z\)
\(\Leftrightarrow a+4\inƯ\left(5\right)\)
\(\Leftrightarrow a+4\in\left\{\pm1;\pm5\right\}\)
+ \(a+4=-1\Leftrightarrow a=-5\)
+ \(a+4=1\Leftrightarrow a=-3\)
+ \(a+4=-5\Leftrightarrow a=-9\)
+ \(a+4=5\Leftrightarrow a=1\)
vậy....
a) Ta có: \(-x+\frac{4}{7}=\frac{1}{3}\)
\(\Leftrightarrow-x=-\frac{5}{21}\)
\(\Rightarrow x=\frac{5}{21}\)
b) Ta có: \(x\div\left(-\frac{1}{3}\right)^2=-\frac{1}{3}\)
\(\Rightarrow x=\left(-\frac{1}{3}\right)^3=-\frac{1}{27}\)
c) \(\left(\frac{3}{5}\right)^5.x=\left(\frac{3}{5}\right)^7\)
\(\Rightarrow x=\left(\frac{3}{5}\right)^2=\frac{9}{25}\)
\(a.-x+\frac{4}{7}=\frac{1}{3}\)
\(-x=\frac{1}{3}-\frac{4}{7} \)
\(-x=\frac{7}{21}-\frac{12}{21}\)
\(-x=\frac{-5}{21}\)
\(x=\frac{5}{21}\)
\(b.x:\left(\frac{-1}{3}\right)^2=\frac{-1}{3}\)
\(x=\frac{-1}{3}.\left(\frac{-1}{3}\right)^2\)
\(x=\frac{-1}{3}.\frac{-1}{3}.\frac{-1}{3}\)
\(x=\frac{-1}{27}\)
\(c.\left(\frac{3}{5}\right)^5.x=\left(\frac{3}{5}\right)^7\)
\(x=\left(\frac{3}{5}\right)^7:\left(\frac{3}{5}\right)^5\)
\(x=\left(\frac{3}{5}\right)^2\)
\(x=\frac{3}{5}.\frac{3}{5}\)
\(x=\frac{9}{25}\)
c)\(a=1+7+7^2+...+7^{100}\)
\(a=\left(1+7\right)+\left(7^2+7^3\right)+...+\left(7^{99}+7^{100}\right)\)
\(a=8+7^2\left(1+7\right)+...+7^{99}\left(1+7\right)\)
\(a=8+7^2.8+...+7^{99}.8\)
\(a=8\left(1+7^2+...+7^{99}\right)\) chia hết cho 8
=> a chia 8 dư 0
a)\(a=\left(1+7+7^2+...+7^{100}\right)\)
=>\(7a=7\left(1+7+7^2+...+7^{100}\right)=7+7^2+7^3+...+7^{101}\)
=>\(7a-a=\left(7+7^2+7^3+...+7^{101}\right)\)\(-\left(1+7+7^2+...+7^{100}\right)\)
=>\(6a=7^{101}-1\Rightarrow a=\frac{7^{101}-1}{6}\)
b)\(6a+1=7^x\Rightarrow6.\frac{7^{101}-1}{6}+1=7^x\Rightarrow7^{101}-1+1=7^x\Rightarrow7^{101}=7^x\)=>x=101