tìm X
[(10-x).2+5]:3-2=3
giúp tôi giải với
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\(x-\dfrac{3}{4}=\dfrac{1}{8}\\ x=\dfrac{1}{8}+\dfrac{3}{4}=\dfrac{7}{8}\)
x - \(\dfrac{3}{4}\)=\(\dfrac{1}{8}\)=>x=\(\dfrac{7}{8}\)
=>5căn x+2-15y=15 và 5căn x+2-2y=71/3
=>-13y=4/3 và căn x+2-3y=3
=>y=-4/39 và căn x+2=3+3y=3-12/39=105/39
=>y=-4/39 và x=887/169
27:(x-3/2)^3=(x-3/2):3
Ta có: \(\dfrac{27}{\left(x-\dfrac{3}{2}\right)^3}=\dfrac{\left(x-\dfrac{3}{2}\right)}{3}\)
\(\Rightarrow\left(x-\dfrac{3}{2}\right)^3.\left(x-\dfrac{3}{2}\right)\)=27.3
\(\Rightarrow\left(x-\dfrac{3}{2}\right)^4\)=81
\(\Rightarrow\left(x-\dfrac{3}{2}\right)^4=3^4\)
\(\Rightarrow\left[{}\begin{matrix}x-\dfrac{3}{2}=4\\x-\dfrac{3}{2}=-4\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=4+\dfrac{3}{2}\\x=-4+\dfrac{3}{2}\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{8}{2}+\dfrac{3}{2}\\x=\dfrac{-8}{2}+\dfrac{3}{2}\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{11}{2}\\x=\dfrac{-5}{2}\end{matrix}\right.\)
Vậy x∈\(\left\{\dfrac{11}{2};\dfrac{-5}{2}\right\}\)
\(\Leftrightarrow x^3-3x^2+3x-1-2x+3x^2-2+6x=-3\)
\(\Leftrightarrow x^3+7x-5=0\)
\(\left(x-5\right)^2=\left(18\dfrac{1}{3}:5\right).\dfrac{11}{3}\)
\(\Leftrightarrow\left(x-5\right)^2=\dfrac{55}{3}.\dfrac{1}{5}.\dfrac{11}{3}\)
\(\Leftrightarrow\left(x-5\right)^2=\dfrac{121}{9}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=\dfrac{11}{3}\\x-5=-\dfrac{11}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{26}{3}\\x=\dfrac{4}{3}\end{matrix}\right.\)
c: Ta có: \(\left(x+1\right)^2\ge0\forall x\)
\(\left(y-\dfrac{1}{3}\right)^2\ge0\forall y\)
Do đó: \(\left(x+1\right)^2+\left(y-\dfrac{1}{3}\right)^2\ge0\forall x,y\)
\(\Leftrightarrow\left(x+1\right)^2+\left(y-\dfrac{1}{3}\right)^2-10\ge-10\forall x,y\)
Dấu '=' xảy ra khi x=-1 và \(y=\dfrac{1}{3}\)
[(10-x).2+5]:3=3+2
\(\Rightarrow\)[(10-x).2+5]:3=5
(10-x).2+5=5.3
(10-x).2+5=15
(10-x).2=15-5
(10-x).2=10
10-x=10:2
10-x=5
\(\Rightarrow\)x=10-5
x=5
x = 5 nho k cho minh nhe