Cứu em
6/5 - x : 1/2 = 1/2
Ai biết giúp em đi em xim cảm ơn ạ
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\(\frac{2}{3}-\frac{1}{5}.\left(\frac{3.x}{2}-\frac{1}{4}\right)=\frac{11}{2}-\frac{1}{4}\)
\(\Leftrightarrow\frac{2}{3}-\frac{1}{5}.\left(\frac{3.x}{2}-\frac{1}{4}\right)=\frac{21}{4}\)
\(\Leftrightarrow\frac{1}{5}.\left(\frac{3.x}{2}-\frac{1}{4}\right)=\frac{2}{3}-\frac{21}{4}\)
\(\Leftrightarrow\frac{1}{5}.\left(\frac{3.x}{2}-\frac{1}{4}\right)=\frac{-55}{12}\)
\(\Leftrightarrow\frac{3.x}{2}-\frac{1}{4}=\frac{-55}{12}:\frac{1}{5}\)
\(\Leftrightarrow\frac{3.x}{2}-\frac{1}{4}=\frac{-275}{12}\)
\(\Leftrightarrow\frac{3.x}{2}=\frac{-275}{12}+\frac{1}{4}\)
\(\Leftrightarrow\frac{3.x}{2}=\frac{-68}{3}\)
\(\Leftrightarrow\left(3.x\right).3=-136\)
\(\Leftrightarrow3.x=-136:3\)
\(\Leftrightarrow3.x=\frac{-136}{3}\)
\(\Leftrightarrow x=\frac{-136}{3}:3\)
\(\Leftrightarrow x=\frac{-136}{9}\)
\(=>x^3=(\sqrt[3]{2\left(\sqrt{3}+1\right)}-\sqrt[3]{2\left(\sqrt{3}-1\right)})^3\)
\(x^3=2\left(\sqrt{3}+1\right)-3.\left[\sqrt[3]{2\left(\sqrt{3}+1\right)}\right]^2.\left[\sqrt[3]{2\left(\sqrt{3}-1\right)}\right]\)
+\(3\left[\sqrt[3]{2\left(\sqrt{3}-1\right)}\right]^2\left[\sqrt[3]{2\left(\sqrt{3}+1\right)}\right]-2\left(\sqrt{3}-1\right)\)
\(x^3=\)
\(4-3\left[\sqrt[3]{2\left(\sqrt{3}+1\right)}\right]\left[\sqrt[3]{2\left(\sqrt{3}-1\right)}\right]\left[\sqrt[3]{2\left(\sqrt{3}+1\right)}-\sqrt[3]{2\left(\sqrt{3}-1\right)}\right]\)
\(x^3=4-3.\left[\sqrt[3]{4\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}\right].\)\(x\)
\(x^3=4-3\left[\sqrt[3]{4\left(3-1\right)}\right].x\)
\(x^3=4-3.2x\)
\(x^3=4-6x\)
thay \(x^3=4-6x\) vào A=>\(A=\left(4-6x+6x-5\right)^{2009}=\left(-1\right)^{2009}=-1\)
\(\left(\frac{2}{3}+x\right)\left(\frac{1}{5}-2x\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}\frac{2}{3}+x=0\\\frac{1}{5}-2x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-\frac{2}{3}\\-2x=-\frac{1}{5}\end{cases}\Leftrightarrow}}\orbr{\begin{cases}x=-\frac{2}{3}\\x=\frac{1}{10}\end{cases}}\)
Vậy:.......
#H
\(m_{H_2O}=1,62\left(g\right)\Rightarrow n_{H_2O}=0,09\left(mol\right)\Rightarrow n_H=0,18\left(mol\right);m_H=0,18.1=0,18\left(g\right)\\ n_{CO_2}=\dfrac{2,64}{44}=0,06\left(mol\right)\Rightarrow n_C=n_{CO_2}=0,06\left(mol\right);m_C=0,06.12=0,72\left(g\right)\\ Vây:m_C+m_H=0,72+0,18=0,9< 1,38\\ \Rightarrow X.có.chứa.O\\ m_O=1,38-0,9=0,48\left(g\right);n_O=\dfrac{0,48}{16}=0,03\left(mol\right)\\ Đặt.X:C_aH_bO_c\left(a,b,c:nguyên,dương\right)\\ Ta.có:a:b:c=0,06:0,18:0,03=2:6:1\\ \Rightarrow CTĐG:C_2H_6O\\ M_X=23.2=46\left(\dfrac{g}{mol}\right)=M_{C_2H_6O}\\ \Rightarrow X:C_2H_6O\)
Em tìm điều kiện xác định của bài toán.
Sau đó bình phương hai vế lên (cả hai vế đều >0) xem ra kết quả gì?
`6/5 - x : 1/2 = 1/2`
`x:1/2=6/5-1/2`
`x:1/2=12/10 - 5/10`
`x:1/2=7/10`
`x=7/10xx1/2`
`x=7/20`
Vậy `x=7/20`
`#LeMichael`
\(\dfrac{6}{5}\) - x : \(\dfrac{1}{2}\) = \(\dfrac{1}{2}\)
x : \(\dfrac{1}{2}\) = \(\dfrac{6}{5}\) - \(\dfrac{1}{2}\)
x : \(\dfrac{1}{2}\) = \(\dfrac{7}{10}\)
x = \(\dfrac{7}{10}\) x \(\dfrac{1}{2}\)
x = \(\dfrac{7}{20}\)