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bài 1 tính giá trị biểu thức
( - 25 ) nhân ( -3 ) nhân x với x = 4
\(\left(-25\right).\left(-3\right).4\)
\(=\left(-25\right).4.\left(-3\right)\)
\(=-100.\left(-3\right)=300\)
( -1 ) nhân ( -4 ) nhân 5 nhân 8 nhân y với y =25
\(\left(-1\right).\left(-4\right).5.8.25\)
\(=4.5.8.25=4.25.5.8\)
\(=100.40=40000\)
( 2ab mũ 2 ) : c với a =4 ; b= -6 ; c =12
\(\left(2.4.\left(-6\right)\right)^2:12\)
\(=\left(-48\right)^2:12\)
\(=2304:12=192\)
[ ( -25 ) nhân ( - 27 ) nhân ( -x ) ] : y với x = 4 ; y = -9
\(\left[\left(-25\right).\left(-27\right).\left(-4\right)\right]:-9\)
\(=-2700:\left(-9\right)\)
\(=300\)
(a mũ 2 _ b mũ 2) : ( a + b ) nhân ( a _ b ) với a + 5 , b = -3
\(\left(5^2-\left(-3\right)^2\right):\left(5-3\right).\left(5+3\right)\)
\(=16:2.8\)
\(=8.8=64\)
1) \(2^x-15=17\)
\(\Leftrightarrow2^x=32=2^5\)
\(\Rightarrow x=5\)
2) \(\left(7x-11\right)^3=25\cdot5^2+200\)
\(\Leftrightarrow\left(7x-11\right)^3=825\)
\(\Leftrightarrow7x-11=\sqrt[3]{825}\)
\(\Leftrightarrow7x=11+\sqrt[3]{825}\)
\(\Rightarrow x=\frac{11+\sqrt[3]{825}}{7}\)
3) \(\left(x+1\right)^{100}-3\left(x+1\right)^{99}=0\)
\(\Leftrightarrow\left(x+1\right)^{99}\left(x-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}\left(x+1\right)^{99}=0\\x-2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-1\\x=2\end{cases}}\)
4) \(4x+5\left(x+3\right)=105\)
\(\Leftrightarrow9x+15=105\)
\(\Leftrightarrow9x=90\)
\(\Rightarrow x=10\)
5) \(5\cdot\left(x-2\right)+10\left(x+3\right)=170\)
\(\Leftrightarrow5\left[x-2+2\left(x+3\right)\right]=170\)
\(\Leftrightarrow3x+4=34\)
\(\Leftrightarrow3x=30\)
\(\Rightarrow x=10\)
a) \(\frac{8}{9}\cdot x-\frac{2}{3}=\frac{1}{3}\cdot x+1\frac{1}{3}\)
=> \(\frac{8x}{9}-\frac{2}{3}=\frac{x}{3}+\frac{4}{3}\)
=> \(\frac{8x}{9}-\frac{6}{9}=\frac{x+4}{3}\)
=> \(\frac{8x-6}{9}=\frac{x+4}{3}\)
=> \(3\left(8x-6\right)=9\left(x+4\right)\)
=> \(24x-18=9x+36\)
=> \(24x-18-9x=36\)
=> \(24x-9x=54\)
=> \(15x=54\)
=> \(5x=18\)
=> \(x=\frac{18}{5}\)
Vậy x = \(\frac{18}{5}\)
b) \(\left(x-\frac{1}{2}\right)\left(\frac{3}{2}-2x\right)=0\)
=> \(\orbr{\begin{cases}x-\frac{1}{2}=0\\\frac{3}{2}-2x=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{1}{2}\\2x=\frac{3}{2}\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{1}{2}\\x=\frac{3}{2}:2=\frac{3}{4}\end{cases}}\)
Vậy \(x\in\left\{\frac{1}{2};\frac{3}{4}\right\}\)
X x \(\dfrac{3}{4}\)+ X x\(\dfrac{1}{5}\)+ X x \(\dfrac{1}{20}\)+ X= 1000
`@` `\text {Ans}`
`\downarrow`
\(\dfrac{2}{3}+\left[\dfrac{4}{5}x-\dfrac{11}{15}\right]=\dfrac{5}{9}\)
`=>`\(\dfrac{4}{5}x-\dfrac{11}{15}=\dfrac{5}{9}-\dfrac{2}{3}\)
`=>`\(\dfrac{4}{5}x-\dfrac{11}{15}=-\dfrac{1}{9}\)
`=>`\(\dfrac{4}{5}x=-\dfrac{1}{9}+\dfrac{11}{15}\)
`=>`\(\dfrac{4}{5}x=\dfrac{28}{45}\)
`=>`\(x=\dfrac{28}{45}\div\dfrac{ 4}{5}\)
`=>`\(x=\dfrac{7}{9}\)
Vậy, `x = 7/9.`
\(\left(\dfrac{3}{4}-3x\right)\cdot\left(1+4x\right)=0\)
`=>`\(\left[{}\begin{matrix}\dfrac{3}{4}-3x=0\\1+4x=0\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}3x=\dfrac{3}{4}\\4x=-1\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=\dfrac{1}{4}\\x=-\dfrac{1}{4}\end{matrix}\right.\)
Vậy, `x \in {1/4; -1/4}.`
\(\left(3:x-1\right)\left(-\dfrac{1}{2}x+5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}3:x-1=0\\-\dfrac{1}{2}x+5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}3:x=1\\-\dfrac{1}{2}x=-5\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=3\\x=10\end{matrix}\right.\)