A = \(10^3-\left\{-5^3.2^3-11.\left[x^2-5.2^3+\left(121-11^2\right)\right]\right\}\)
Tính A
K
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1
10 tháng 4 2019
\(A=1500-\left\{5^3.2^3-11.\left[7^2-5.2^3+8\left(11^2-121\right)\right]\right\}\)
\(A=1500-\left\{125.8-11.\left[49-5.8+8\left(121-121\right)\right]\right\}\)
\(A=1500-\left\{1000-11\left[49-40+8.0\right]\right\}\)
\(A=1500-\left\{1000-11.9\right\}\)
\(A=1500-\left\{1000-99\right\}\)
\(A=1500-901=599\)
C
23 tháng 2 2017
Ta tính các số mũ thành số hết
\(A=-1500-\left\{125.8-11.\left[49-40+8\left(121-121\right)\right]\right\}.2\)
\(A=-1500-\left\{1000-11.\left(9+0\right)\right\}.2\)
\(A=-1500-\left(1000-99\right).2\)
\(A=-1500-901.2\)
\(A=-1500-1802=-3302\)
25 tháng 3 2017
\(1000-\left\{\left(-5\right)^3.\left(-2\right)^3-11.[7^2-5.2^3+8.\left(11^2-121\right)]\right\}\)
=\(1000-\left\{[\left(-5\right).\left(-2\right)]^3-11.[7^2-5.2^3+2^3.\left(11^2-11^2\right)]\right\}\)
= \(1000-\left\{1000-11.[7^2-2^3.\left(5+0\right)]\right\}\)
= \(1000-[1000-11.\left(7^2-2^3.5\right)\)
= \(1000-[1000-11.\left(49-40\right)]\)
= \(1000-\left(1000-11.9\right)\)
= \(1000-\left(1000-99\right)=1000-1000+99\)
= 0 + 99 = 99
25 tháng 10 2017
\(B=\frac{2^{19}.27^3+15.4^9.9^4}{6^9.2^{10}+12^{10}}\)
\(B=\frac{2^{19}.3^9+3.5.2^{18}.3^8}{2^9.3^9.2^{10}+3^{10}.2^{20}}\)
\(B=\frac{2^{19}.3^9+3^9.5.2^{18}}{2^{19}.3^9+3^{10}.2^{20}}\)
\(B=\frac{2^{18}.3^9.\left(2+5\right)}{2^{19}.3^9\left(1+3.2\right)}\)
\(B=\frac{7}{2.7}\)
\(B=\frac{1}{2}\)
\(C=\frac{2^{13}.4^{11}-16^9}{\left(3.2^{17}\right)^2}\)
\(C=\frac{2^{13}.2^{22}-2^{36}}{3^2.2^{34}}\)
\(C=\frac{2^{35}-2^{36}}{3^2.2^{34}}\)
\(C=\frac{2^{35}\left(1-2\right)}{3^2.2^{34}}\)
\(C=\frac{-2}{9}\)
\(D=\frac{4^7.2^8}{3.2^{15}.16^2-5.2^2.\left(2^{10}\right)^2}\)
\(D=\frac{2^{14}.2^8}{3.2^{15}.2^8-5.2^2.2^{20}}\)
\(D=\frac{2^{14}.2^8}{3.2^{23}-5.2^{22}}\)
\(D=\frac{2^{22}}{2^{22}\left(3.2-5\right)}\)
\(D=1\)
Y
11 tháng 1 2023
\(8,1-\left(x-6\right)=4\left(2-2x\right)\)
\(\Leftrightarrow1-x+6=8-8x\)
\(\Leftrightarrow-x+8x=8-1-6\)
\(\Leftrightarrow7x=1\)
\(\Leftrightarrow x=\dfrac{1}{7}\)
\(9,\left(3x-2\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-2=0\\x+5=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-5\end{matrix}\right.\)
\(10,\left(x+3\right)\left(x^2+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x^2+2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=\varnothing\end{matrix}\right.\)
2
11 tháng 1 2023
`8)1-(x-5)=4(2-2x)`
`<=>1-x+5=8-6x`
`<=>5x=2<=>x=2/5`
`9)(3x-2)(x+5)=0`
`<=>[(x=2/3),(x=-5):}`
`10)(x+3)(x^2+2)=0`
Mà `x^2+2 > 0 AA x`
`=>x+3=0`
`<=>x=-3`
`11)(5x-1)(x^2-9)=0`
`<=>(5x-1)(x-3)(x+3)=0`
`<=>[(x=1/5),(x=3),(x=-3):}`
`12)x(x-3)+3(x-3)=0`
`<=>(x-3)(x+3)=0`
`<=>[(x=3),(x=-3):}`
`13)x(x-5)-4x+20=0`
`<=>x(x-5)-4(x-5)=0`
`<=>(x-5)(x-4)=0`
`<=>[(x=5),(x=4):}`
`14)x^2+4x-5=0`
`<=>x^2+5x-x-5=0`
`<=>(x+5)(x-1)=0`
`<=>[(x=-5),(x=1):}`
\(A=10^3-\left\{-5^3.2^3-11.\left[x^2-5.2^3+\left(121-11^2\right)\right]\right\}\)
\(A=10^3-\left\{\left(-10\right)^3-11.\left[x^2-40\right]\right\}\)
\(A=10^3-\left\{\left(-10\right)^3-11x^2+440\right\}\)
\(A=10^3+10^3+11x^2-440\)
\(A=2000-440+11x^2\)
\(A=1560+11x^2\)