Là 30 cm

Tick nha :)

a) \(\left(x-y\right)^3+\left(y-z\right)^3+\left(z-x\right)^3\)

\(=\left[\left(x-y\right)^3+\left(y-z\right)^3\right]+\left(z-x\right)^3\)

\(=\left[\left(x-y\right)+\left(y-z\right)\right]\left[\left(x-y\right)^2-\left(x-y\right)\left(y-z\right)+\left(y-z\right)^2\right]+\left(z-x\right)^3\)

\(=\left(x-z\right)\left[\left(x-y\right)^2-\left(x-y\right)\left(y-z\right)+\left(y-z\right)^2\right]-\left(x-z\right)^3\)

\(=\left(x-z\right)\left[\left(x-y\right)^2-\left(x-y\right)\left(y-z\right)+\left(y-z\right)^2-\left(x-z\right)^2\right]\)

\(=\left(x-z\right)\left\{\left[\left(x-y\right)^2-\left(x-y\right)\left(y-z\right)\right]+\left[\left(y-z\right)^2-\left(x-z\right)^2\right]\right\}\)

\(=\left(x-z\right)\left[\left(x-y\right)\left(x-y-y+z\right)+\left(y-z-x+z\right)\left(y-z+x-z\right)\right]\)

\(=\left(x-z\right)\left[\left(x-y\right)\left(x-2y+z\right)+\left(y-x\right)\left(y-2z+x\right)\right]\)

\(=\left(x-z\right)\left[\left(x-y\right)\left(x-2y+z\right)-\left(x-y\right)\left(y-2z+x\right)\right]\)

\(=\left(x-z\right)\left[\left(x-y\right)\left(x-2y+z-y+2z-x\right)\right]\)

\(=\left(x-z\right)\left[\left(x-y\right)\left(x-2y+z-y+2z-x\right)\right]\)

\(=\left(x-z\right)\left[\left(x-y\right)\left(3z-3y\right)\right]\)

\(=3\left(x-z\right)\left(x-y\right)\left(z-y\right)\)

b) \(P=x^2-x-5\)

\(\Leftrightarrow P=x^2-x+\dfrac{1}{4}-\dfrac{21}{4}\)

\(\Leftrightarrow P=\left(x^2-2.x.\dfrac{1}{2}+\dfrac{1}{4}\right)-\dfrac{21}{4}\)

\(\Leftrightarrow P=\left(x-\dfrac{1}{2}\right)^2-\dfrac{21}{4}\)

Vì \(\left(x-\dfrac{1}{2}\right)^2\ge0\)

Nên \(\left(x-\dfrac{1}{2}\right)^2-\dfrac{21}{4}\ge\dfrac{-21}{4}\)

Vậy GTNN của \(P=\dfrac{-21}{4}\) khi \(x-\dfrac{1}{2}=0\Leftrightarrow x=\dfrac{1}{2}\)

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

\(\Leftrightarrow\left(x-1\right)\left(x+3\right)=\left(x+2\right)\left(x-2\right)\)

\(\Leftrightarrow x^2+3x-x-3=x^2-4\)

\(\Leftrightarrow x^2+3x-x-3-\left(x^2-4\right)=0\)

\(\Leftrightarrow x^2+3x-x-3-x^2+4=0\)

\(\Leftrightarrow2x+1=0\)

\(\Leftrightarrow2x=-1\)

\(\Leftrightarrow x=\dfrac{-1}{2}\)

Vậy \(x=\dfrac{-1}{2}\)

a) \(x^2-x-6=0\)

\(\Leftrightarrow x^2-3x+2x-6=0\)

\(\Leftrightarrow\left(x^2-3x\right)+\left(2x-6\right)=0\)

\(\Leftrightarrow x\left(x-3\right)+2\left(x-3\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x+2=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)

Vậy..........

a) \(M=\left(x-3\right)^3-\left(x+1\right)^3+12x\left(x-1\right)\)

\(\Leftrightarrow M=\left(x^3-9x^2+27x-27\right)-\left(x^3+3x^2+3x+1\right)+\left(12x^2-12x\right)\)

\(\Leftrightarrow M=x^3-9x^2+27x-27-x^3-3x^2-3x-1+12x^2-12x\)

\(\Leftrightarrow M=12x-28\)

b) Thay \(x=\dfrac{-2}{3}\) vào biểu thức M ta được:

\(12.\left(\dfrac{-2}{3}\right)-28=-8-28=-36\)

Vậy giá trị của biểu thức M tại \(x=\dfrac{-2}{3}\) là -36

c) \(12x-28=-16\)

\(\Leftrightarrow12x=-16+28\)

\(\Leftrightarrow12x=12\)

\(\Leftrightarrow x=1\)

Vậy x=1 để \(M=-16\)

minh khong biet cach giai vi minh moi hoc lop 5

a) \(\dfrac{2}{x+3}+\dfrac{1}{x}\) [ MTC: x(x+3) ]

\(=\dfrac{x.2}{x\left(x+3\right)}+\dfrac{1\left(x+3\right)}{x\left(x+3\right)}\)

\(=\dfrac{2x+x+3}{x\left(x+3\right)}\)

\(=\dfrac{3x+3}{x\left(x+3\right)}\)

\(=\dfrac{3\left(x+1\right)}{x\left(x+3\right)}\)

b) \(\dfrac{x+1}{2x-2}+\dfrac{-2x}{x^2-1}\)

\(=\dfrac{x+1}{2\left(x-1\right)}+\dfrac{-2x}{\left(x-1\right)\left(x+1\right)}\) \(\left[MTC:2\left(x-1\right)\left(x+1\right)\right]\)

\(=\dfrac{\left(x+1\right)\left(x+1\right)}{2\left(x-1\right)\left(x+1\right)}+\dfrac{-2x.2}{2\left(x-1\right)\left(x+1\right)}\)

\(=\dfrac{\left(x+1\right)^2-4x}{2\left(x-1\right)\left(x+1\right)}\)

\(=\dfrac{\left(x^2+2x+1\right)-4x}{2\left(x-1\right)\left(x+1\right)}\)

\(=\dfrac{\left(x^2-2x+1\right)}{2\left(x-1\right)\left(x+1\right)}\)

\(=\dfrac{\left(x-1\right)^2}{2\left(x-1\right)\left(x+1\right)}\)

\(=\dfrac{\left(x-1\right)}{2\left(x+1\right)}\)

\(2x=3y=5z\Rightarrow\dfrac{x}{\dfrac{1}{2}}=\dfrac{y}{\dfrac{1}{3}}=\dfrac{z}{\dfrac{1}{5}}\) và \(x+y-z=-66\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\dfrac{x}{\dfrac{1}{2}}=\dfrac{y}{\dfrac{1}{3}}=\dfrac{z}{\dfrac{1}{5}}=\dfrac{x+y-z}{\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{5}}=\dfrac{-66}{\dfrac{19}{30}}=\dfrac{-1980}{19}\)

\(\dfrac{x}{\dfrac{1}{2}}=\dfrac{-1980}{19}\Rightarrow x=\dfrac{-1980}{19}.\dfrac{1}{2}=\dfrac{-990}{19}\)

\(\dfrac{y}{\dfrac{1}{3}}=\dfrac{-1980}{19}\Rightarrow y=\dfrac{-1980}{19}.\dfrac{1}{3}=\dfrac{-660}{19}\)

\(\dfrac{z}{\dfrac{1}{5}}=\dfrac{-1980}{19}\Rightarrow z=\dfrac{-1980}{19}.\dfrac{1}{5}=\dfrac{-396}{19}\)

Vậy................

a) \(100:\left\{250:\left[450-\left(4.5^3-25.4\right)\right]\right\}\)

\(=100:\left\{250:\left[450-\left(4.125-25.4\right)\right]\right\}\)

\(=100:\left\{250:\left[450-\left(500-100\right)\right]\right\}\)

\(=100:\left[250:\left(450-400\right)\right]\)

\(=100:\left(250:50\right)\)

\(=100:5\)

\(=20\)

b) \(4\left(18-15\right)-\left(5-3\right).3^2\)

\(=4.3-2.3^2\)

\(=4.3-2.9\)

\(=12-18\)

​\(=-6\)​

a) \(100:\left\{250:\left[450-\left(4.5^3-2^2.25\right)\right]\right\}\)

\(=100:\left\{250:\left[450-\left(4.125-4.25\right)\right]\right\}\)

\(=100:\left\{250:\left[450-\left(500-100\right)\right]\right\}\)

\(=100:\left[250:\left(450-400\right)\right]\)

\(=100:\left(250:50\right)\)

\(=100:5\)

\(=20\)

b) \(109.5^2-3^2.25\)

\(=109.25-9.25\)

\(=25\left(109-9\right)\)

\(=25.100\)

\(=2500\)

c) \(\left[5^2.6-20.\left(37-2^5\right)\right]:10-20\)

\(=\left[5^2.6-20.\left(37-32\right)\right]:10-20\)

\(=\left(5^2.6-20.5\right):10-20\)

\(=\left(25.6-20.5\right):10-20\)

\(=\left(150-100\right):10-20\)

\(=50:10-20\)

\(=5-20\)

\(=-15\)