611 - [ 2 . ( x2 - 13 ) + 116 ] = 449
Tìm x
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\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\Rightarrow\dfrac{x^2}{4}=\dfrac{y^2}{9}=\dfrac{z^2}{16}\)
Áp dụng t/c dtsbn:
\(\dfrac{x^2}{4}=\dfrac{y^2}{9}=\dfrac{z^2}{16}=\dfrac{x^2+y^2+z^2}{4+9+16}=\dfrac{116}{29}=4\)
\(\Rightarrow\left\{{}\begin{matrix}x^2=4.4=16\\y^2=4.9=36\\z^2=16.16=16^2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=4\\y=6\\z=16\end{matrix}\right.\\\left\{{}\begin{matrix}x=-4\\y=-6\\z=-16\end{matrix}\right.\end{matrix}\right.\)
\(\frac{30}{13}\)× \(\frac{52}{29}\)×\(\frac{116}{240}\)= \(\frac{3×10×13×4×29×4}{13×29×2×3×4×10}\)( gạch số giống nhau đi )=\(\frac{4}{2}\)= 2
a) A= 100 - 116 : 4
=100-29
=71
b) B= 100 - [ 116 - (13 - 5)2]
=100-[116-82]
=100-(116-64)
=100-52
=48
c) C= |- 65| + (- 42) + |65|
= 65+(-42)+65
= 88
5) Ta có: \(\dfrac{\left(5\sqrt{3}+\sqrt{50}\right)\left(5-\sqrt{24}\right)}{\sqrt{75}-5\sqrt{2}}\)
\(=\dfrac{5\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)^2}{5\left(\sqrt{3}-\sqrt{2}\right)}\)
=1
\(100-\left(116-\left(13-5^2\right)\right)=100-\left(116-\left(-12\right)\right)\)
\(=100-\left(116+12\right)=100-128=-28\)
100-[116-(13-52)]
=100-[116-(13-25)]
=100-(116-13+25)
=100-(103+25)
=100-128
=-28
\(100-\left[116-\left(13-5\right)^2\right]\)
\(=100-\left(116-8^2\right)\)
\(=100-\left(116-64\right)\)
\(=100-52\)
\(=48\)
\(611-\left[2.\left(x^2-13\right)+116\right]=449.\)
\(\Rightarrow2.\left(x^2-13\right)+116=611-449\)
\(\Rightarrow2.\left(x^2-13\right)+116=162\)
\(\Rightarrow2.\left(x^2-13\right)=162-116\)
\(\Rightarrow2.\left(x^2-13\right)=46\)
\(\Rightarrow x^2-13=46:2\)
\(\Rightarrow x^2-13=23\)
\(\Rightarrow x^2=23+13\)
\(\Rightarrow x^2=36\)
\(\Rightarrow x=\orbr{\begin{cases}6\\-6\end{cases}}\)
611 - [ 2 * ( x2 - 13 ) + 116 ] = 449
2(x2 - 13 ) +116 = 162
2 * ( x2 - 13 ) = 162 - 116
2 * ( x2 - 13 ) = 46
x2 - 13 = 23
x2 = 36
x = 6
Vậy x = 6