Tìm x, y

Ta có: \(\hept{\begin{cases}\left(x-2\right)^{2012}\ge0\forall x\inℤ\\\left|y^2-9\right|^{2014}\ge0\forall x\inℤ\end{cases}}\)

\(\left(x-2\right)^{2012}+\left|y^2-9\right|^{2014}=0\)

\(\Rightarrow\hept{\begin{cases}\left(x-2\right)^{2012}=0\\\left|y^2-9\right|^{2014}=0\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x-2=0\\y^2-9=0\end{cases}}\Rightarrow\hept{\begin{cases}x=2\\y^2=9\end{cases}}\Rightarrow\hept{\begin{cases}x=2\\y^2=\pm3\end{cases}}\)

Vậy \(x=2\) và \(y=\pm3\).

Sửa lại phần ta có: \(\hept{\begin{cases}\left(x-2\right)^{2012}\ge0\forall x\inℚ\\\left|y^2-9\right|^{2014}\ge0\forall y\inℚ\end{cases}}\)

\(B=\frac{4^2.25^2+32.125}{2^3.5^2}\)

\(=\frac{\left(2^2\right)^2.\left(5^2\right)^2+2^5.5^3}{2^3.5^2}\)

\(=\frac{2^4.5^4+2^5.5^3}{2^3.5^2}\)

\(=\frac{2^3.5^2.\left(2.5^2+2^2.5\right)}{2^3.5^2}\)

\(=2.5^2+2^2.5\)

\(=2.25+4.5\)

\(=50+20\)

\(=70\)

Bài làm

\(B=\frac{4^2\cdot25^2+32\cdot125}{2^3\cdot5^2}\)

\(B=\frac{\left(2^2\right)^2\cdot\left(5^2\right)^2+2^5\cdot5^3}{2^3\cdot5^2}\)

\(B=\frac{2^4\cdot5^2+2^5\cdot5^3}{2^3\cdot5^2}\)

\(B=\frac{2^4\left(5^2+2\cdot5^3\right)}{2^3.5^2}\)

\(B=\frac{2^4\left[5^2\left(1+2\cdot5\right)\right]}{2^3.5^2}\)

\(B=\frac{2^4\cdot5^2\cdot11}{2^3\cdot5^2}\)

\(B=2.11=22\)

Vậy B = 22

x = -9/10,

x = -1/10

\(\left(2x+1\right)^2=\frac{16}{25}\)

\(\Rightarrow\left(2x+1\right)^2=\left(\frac{4}{5}\right)^2\)

\(\Leftrightarrow2x+1=\frac{4}{5}\)

\(\Leftrightarrow2x=-\frac{1}{5}\)

\(\Leftrightarrow x=-\frac{1}{10}\)

Bài làm:

Ta có: \(\left[\left(2x-0,3\right)-2\right]=8\)

\(\Leftrightarrow2x-0,3=10\)

\(\Leftrightarrow2x=10,3\)

\(\Leftrightarrow x=5,15\)

\(\left\{\left[2x-0,3\right]-2\right\}=8\)

= \(\left[2x-0,3\right]=10\)

= 2x = 10,3

x = 5,15

\(3x^2-2x=0\)

=> \(x.\left(3x-2\right)=0\)

=> x = 0 hoặc 3x - 2 =0

 x = 0               3x = 2

x = 0                x = \(\frac{2}{3}\)

vậy x = 0 hoặc x = \(\frac{2}{3}\)

\(3x^2-2x=0\)

\(\Delta'=1\)

\(x_1=\frac{1+1}{3}=\frac{2}{3}\)

\(x_2=\frac{1-1}{3}=0\)

\(\frac{x-1}{x+2}=\frac{2}{3}\)\(\Leftrightarrow3\left(x-1\right)=2\left(x+2\right)\)

\(\Leftrightarrow3x-3=2x+4\)

\(\Leftrightarrow3x-2x=4+3\)

\(\Leftrightarrow x=7\)

Vậy \(x=7\)

\(\frac{x-1}{x+2}=\frac{2}{3}\)

\(\Leftrightarrow3\left(x-1\right)=2\left(x+2\right)\)

\(\Leftrightarrow3x-3=2x+4\)

\(\Leftrightarrow3x-2x=3+4\)

\(\Leftrightarrow x=7\)

vẽ AE _|_ CD tại E, gọi M là giao điểm của AE và CK

\(\Delta\)ADC có CK,AE ;à hai đường cao cắt nhau tại M

=> M là trực tâm tam giác ADC

=> DM_|_AC, AB _|_AC => AB//DM(đpcm)

\(\Delta\)ADB=\(\Delta\)DAM (g.c.g) => AB=DM

\(\Delta\)HAB=\(\Delta\)KDM (cạnh huyền-góc nhọn) => AH=DK (đpcm)

Bài giải : a) Ta có : góc XAB = ( góc ABC + góc ACB ) => 1/2 góc BAX = 1/2 ( góc ABC + góc ACB ) 

                   => góc EAB = 1/2 ( góc B + góc C ) = B+ C/2 .

b) Ta có : góc B + góc C = 180⁰ - 60⁰ = 120⁰ => góc EAB = 1/2.120 = 60⁰. Xét tam giác AEC ta lại có : góc C = 180⁰ - góc EAC - góc AEC = 180⁰ - ( góc EAB + góc ABC ) - góc CEA = 180⁰ - ( 60⁰ + 60⁰ ) - 15⁰ = 45⁰. Xét tam giác ABC : góc A + góc B+ góc C = 180⁰ 

=> góc B = 180⁰ - góc A - góc C = 180⁰ - 60⁰ -45⁰ = 75⁰ .

\(\frac{1}{x.\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}-\frac{1}{x}=\frac{1}{2010}\).

\(\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x}=\frac{1}{2010}\)

\(=-\frac{1}{x+3}=\frac{1}{2010}\)

\(x=2010-\left(-3\right)=2013\)

\(\frac{x-2}{-\frac{2}{9}}=\frac{-2}{x-2}\)

=> (x - 2)² = \(\frac{-2}{9}.\left(-2\right)\)

=> (x - 2)² = 9

=> \(\orbr{\begin{cases}x-2=3\\x-2=-3\end{cases}}\Rightarrow\orbr{\begin{cases}x=5\\x=-1\end{cases}}\)

\(\frac{x-2}{\frac{-2}{9}}=\frac{-2}{x-2}\)

\(\Rightarrow\left(x-2\right).\left(x-2\right)=\frac{-2}{9}.\left(-2\right)\)

\(\Rightarrow\left(x-2\right)^2=\frac{4}{9}\)

\(\Rightarrow\left(x-2\right)^2=\left(\frac{2}{3}\right)^2\)

\(\Rightarrow\orbr{\begin{cases}x-2=\frac{2}{3}\\x-2=-\frac{2}{3}\end{cases}}\) \(\Rightarrow\orbr{\begin{cases}x=\frac{2}{3}+2\\x=-\frac{2}{3}+2\end{cases}}\) \(\Rightarrow\orbr{\begin{cases}x=\frac{8}{3}\\x=\frac{4}{3}\end{cases}}\)

Vậy \(x=\frac{8}{3}\) hoặc \(x=\frac{4}{3}\)

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