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123 x 123 x 123 x 123 x 123 x 123 : 123567 = 28023873,62
úm ba la xin tích
123 x 123 x 123 x 123 x 123 x 123 : 1234567
= 1236 : 1234567
= 2804891,101
11 x 11 = ( 1 + 1 ) x ( 1 + 1 ) = 2 x 2 = 4
22 x 22 = ( 2 + 2 ) x ( 2 + 2 ) = 4 x 4 = 16
33 x 33 = ( 3 + 3 ) x ( 3 + 3 ) = 6 x 6 = 36
Đáp án 1 : 36
11 x11 = ( 1 + 1 ) x ( 1 + 1 ) = 2 x 2 = 4
22 x 22 = ( 2 + 2 ) x ( 2 + 2 ) = 4 x 4 = 16
33 x 33 = ( 3 + 3 ) x ( 3 + 3 ) = 6 x 6 =36
Đáp án 2 : 28
11 x 11 = 11 + ( 1 + 1 ) - 9 = 4
22 x 22 = 22 + ( 2 + 2 ) - 10 = 16
33 x 33 = 33 + ( 3 + 3 ) - 11 = 28
Còn nhiều đáp án khác nữa nhé
\(\dfrac{-7}{11}=\dfrac{x}{22}\Rightarrow\dfrac{-14}{22}=\dfrac{x}{22}\Rightarrow x=-14\)
\(\dfrac{-5}{11}=\dfrac{x}{33}\Rightarrow\dfrac{-15}{33}=\dfrac{x}{33}\Rightarrow x=-15\)
\(\dfrac{-7}{11}\)=\(\dfrac{x}{22}\)\(\Rightarrow\)11x=-7.22\(\Rightarrow\)11x=-154\(\Rightarrow\)x=-154:11=-14
\(\dfrac{-5}{11}\)=\(\dfrac{x}{33}\)\(\Rightarrow\)11x=-5.33\(\Rightarrow\)11x=-165\(\Rightarrow\)x=-165:11=-15
A = ( 15/22 - 2/22 ) : 1/33 - ( 6/84 - 8/84) : 1/35 + 1 + 1/5 . ( 3/12 - 2/12 - 12/12 ) . 5/11
A = 13/22 . 33 - (-1/42) . 35 + 1 + 1/5 . - 11/12 . 5/11
A = 39/2 - ( -5/6 ) + 1 + - 11/60 .5/11
A = 39/2 + 5/6 + 1 + (- 1/12)
A = 234/12 + 10 /12 + 12/12 + (-1/12)
A = 255/12
\(A=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{99}}\)
\(\Rightarrow\dfrac{A}{3}=\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{100}}\)
\(\Rightarrow A-\dfrac{A}{3}=\dfrac{2A}{3}=\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}}\right)-\left(\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{100}}\right)\)
\(\Rightarrow\dfrac{2A}{3}=\left(\dfrac{1}{3^2}-\dfrac{1}{3^2}\right)+\left(\dfrac{1}{3^3}-\dfrac{1}{3^3}\right)+...+\left(\dfrac{1}{3^{99}}-\dfrac{1}{3^{99}}\right)+\left(\dfrac{1}{3}-\dfrac{1}{3^{100}}\right)=\dfrac{1}{3}-\dfrac{1}{3^{100}}\)
\(\Rightarrow2A=3\cdot\left(\dfrac{1}{3}-\dfrac{1}{3^{100}}\right)\)
\(\Rightarrow\text{A}=\dfrac{1-\dfrac{1}{3^{99}}}{2}\)
\(\Rightarrow A=\dfrac{1}{2}-\dfrac{1}{2.3^{99}}< \dfrac{1}{2}\)
5x + 11 = 9 - x
<=> 5x = 9 - 11 - x
<=> 5x + x = - 20
<=> 6x = - 20
=> x =\(\frac{-20}{6}\)
\(2^{x+22}-4^{x+11}=0\)
\(\Rightarrow2^{x+22}-2^{2x+22}=0\)
\(\Rightarrow2^x\cdot\left(2^{22}-2^{x+22}\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}2^x=0\left(L\right)\\2^{22}-2^{x+22}=0\end{matrix}\right.\)
\(\Rightarrow2^{x+22}=2^{22}\)
\(\Rightarrow x+22=22\)
\(\Rightarrow x=22-22\)
\(\Rightarrow x=0\)
Vậy x=0
\(2^{x+22}-4^{x+11}\text{=}0\)
\(2^{x+22}\text{=}4^{x+11}\)
\(2^x.2^{22}\text{=}4^x.4^{11}\)
\(2^x.2^{22}\text{=}4^x.\left(2^2\right)^{11}\)
\(2^x.2^{22}\text{=}4^x.2^{22}\)
\(2^x\text{=}4^x\)
\(x\text{=}0\)
1)
x^3 -16x=0`
`<=>x(x^2 -16)=0`
\(< =>\left[{}\begin{matrix}x=0\\x^2-16=0\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=0\\x^2=16\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=0\\x=4\\x=-4\end{matrix}\right.\)
b)
`x^4 -2x^3=0`
`<=>x^3 (x-2)=0`
\(< =>\left[{}\begin{matrix}x^3=0\\x-2=0\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)
3)
`(2x-11)(x^2 -1)=0`
\(< =>\left[{}\begin{matrix}2x-11=0\\x^2-1=0\end{matrix}\right.\\ < =>\left[{}\begin{matrix}2x=11\\x^2=1\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=\dfrac{11}{2}\\x=1\\x=-1\end{matrix}\right.\)
4)
`x^3 -36x=0`
`<=>x(x^2 -36)=0`
\(< =>\left[{}\begin{matrix}x=0\\x^2-36=0\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=0\\x^2=36\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=0\\x=6\\x=-6\end{matrix}\right.\)
5)
`2x+19=0`
`<=>2x=-19`
`<=>x=-19/2`
11 + 22 + ( \(x\) x 2 + \(x\) x 199 ) : 33 = 123 x 123
33 + \(x\) ( 2 + 199) : 33 = 15129
201\(x\) : 33 = 15129 - 33
201\(x\) : 33 = 15096
201\(x\) = 15096 x 33
201\(x\) = 498168
\(x\) = 498168 : 201
\(x\) = 166056/67