(2/3 * x )+3/4 = 3
giúp nha
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a: =5/6(1/8+2/3)=5/6*19/24=95/144
b: =3/4(7/5-1/2)=27/40
c: =35/24*2/3=35/36
\(x-\dfrac{3}{4}=\dfrac{1}{8}\\ x=\dfrac{1}{8}+\dfrac{3}{4}=\dfrac{7}{8}\)
x - \(\dfrac{3}{4}\)=\(\dfrac{1}{8}\)=>x=\(\dfrac{7}{8}\)
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∈\(\left\{\dfrac{11}{2};\dfrac{-5}{2}\right\}\)
hơi thiếu nha bạn , nên mình ko hiểu cho lắm ý. Bạn thông cảm nha.
Bài 1:
a) \(x\left(x+1\right)+x\left(x-1\right)-2x^2\)
\(=x^2+x+x^2-x-2x^2\)
\(=2x^2-2x^2\)
\(=0\)
b) \(\left(x+2\right)\left(x^2-x+1\right)-\left(x-2\right)\left(x^2+x+1\right)\)
\(=x^3-x^2+x+2x^2-2x+2-x^3-x^2-x+2x^2+2x+2\)
\(=\left(x^3-x^3\right)+\left(-x^2+2x^2-x^2+2x^2\right)+\left(x-2x-x+2x\right)+\left(2+2\right)\)
\(=2x^2+4\)
c) \(\left(3-x\right)^2+2\left(x-3\right)\left(x+7\right)+\left(x+7\right)^2\)
\(=\left(x-3\right)^2+2\left(x-3\right)\left(x+7\right)+\left(x+7\right)^2\)
\(=\left[\left(x-3\right)+\left(x+7\right)\right]^2\)
\(=\left(x-3+x+7\right)^2\)
\(=\left(2x+4\right)^2\)
\(\Leftrightarrow x^3-3x^2+3x-1-2x+3x^2-2+6x=-3\)
\(\Leftrightarrow x^3+7x-5=0\)
\(\frac{2x}{3}+\frac{3}{4}=3\)
\(\frac{2x}{3}=3-\frac{3}{4}\)
\(\frac{2x}{3}=\frac{9}{4}\)
\(x=\frac{27}{8}\)
TK VS KB NHA!
(2/3 * x )+3/4 = 3
= 2/3 * x = 3 - 3/4 = 9/4
2/3 * x = 9/4
x = 9/4 : 2/3 = 27/8