\(\left(x^3-x^2\right)-4x^2+8x-4=0\)

\(\Leftrightarrow x^3-x^2-4x^2+8x-4=0\)

\(\Leftrightarrow x^3-5x^2+8x-4=0\)

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

\(\Leftrightarrow x^2\left(x-1\right)-4x\left(x-1\right)+4\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^2-4x+4\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\\left(x-2\right)^2=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=1\\x-2=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=1\\x=2\end{cases}}\)

Vậy ...

\(\left(x^3-x^2\right)-4x^2+8x-4=0\)

\(\Leftrightarrow x^3-x^2-4x^2+8x-4=0\)

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

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

\(\Leftrightarrow x^2\left(x-1\right)-4x\left(x-1\right)+4\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^2-4x+4\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\x-2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\\x=2\end{cases}}}\)

9a² + b² + c² - 18a + 2c - 6b + 19 = 0

<=> (9a² - 18a + 9) + (b² - 6b + 9)  + (c² + 2c + 1) = 0

<=> (3a - 3)² + (b - 3)² + (c + 1)² = 0

<=> \(\hept{\begin{cases}3a-3=0\\b-3=0\\c+1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}a=1\\b=3\\c=-1\end{cases}}\)

Vậy a = 1 ; b = 3 ; c = -1 là nghiệm phương trình

\(9a^2+b^2+c^2-18a+2c-6b+19=0\)

\(\Leftrightarrow\left(9a^2-18a+9\right)+\left(b^2-6b+9\right)+\left(c^2+2c+1\right)=0\)

\(\Leftrightarrow\left(3a-3\right)^2+\left(b-3\right)^2+\left(c+1\right)^2=0\)

Vì \(\hept{\begin{cases}\left(3a-3\right)^2\\\left(b-3\right)^2\\\left(c+1\right)^2\end{cases}\ge0\forall a,b,c}\)

\(\Rightarrow\left(3a-3\right)^2+\left(b-3\right)^2+\left(c+1\right)^2\ge0\forall a,b,c\)

Dấu bằng xảy ra khi \(\hept{\begin{cases}3a-3=0\\b-3=0\\c+1=0\end{cases}\Leftrightarrow\hept{\begin{cases}a=1\\b=3\\c=-1\end{cases}}}\)

Vậy...

Tổng của chiều dài và chiều rộng là : 

            100 : 2 = 50 (m)

Trung bình cộng của chiều dài và rộng là :

           50 : 2 = 25 (m) 

Vì chiều dà hơn chiều rộng 3m nên chiều dài trừ đi chiều rộng thì bằng 3m

Chiều dài là :

          (50+3):2=26,5(m)

Chiều rộng là :

          26,5-3=23,5 (m)

                 Đáp số : 23,5 m

đáp số là 23,5m

AM - GM cho 4 số ta được : 

\(a^4+b^4+c^4+d^4\ge4\sqrt[4]{a^4b^4c^4d^4}=4abcd\)( đpcm )

\(a^4+b^4+c^4+d^4\)

\(=\left(a^4+b^4\right)+\left(c^4+d^4\right)\)

\(=\left[\left(a^2\right)^2+\left(b^2\right)^2\right]+\left[\left(c^2\right)^2+\left(d^2\right)^2\right]\ge2a^2b^2+2c^2d^2\)

\(=2\left[\left(ab\right)^2+\left(cd\right)^2\right]\ge2.2abcd\ge4abcd\)

Dấu"=" xảy ra khi \(a=b=c=d\)

Nguồn:hoidap247

Ta có:\(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}=1\)

\(\Leftrightarrow\left(a+b+c\right)\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right)=a+b+c\)

\(\Leftrightarrow\frac{a\left(a+b+c\right)}{b+c}+\frac{b\left(a+b+c\right)}{c+a}+\frac{c\left(a+b+c\right)}{a+b}=a+b+c\)

\(\Leftrightarrow\frac{a^2+a\left(b+c\right)}{b+c}+\frac{b^2+b\left(c+a\right)}{c+a}+\frac{c^2+c\left(a+b\right)}{a+b}=a+b+c\)

\(\Leftrightarrow a+\frac{a^2}{b+c}+b+\frac{b^2}{c+a}+c+\frac{c^2}{a+b}=a+b+c\)

\(\Leftrightarrow\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}=a+b+c-a-b-c\)

\(\Leftrightarrow\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}=0\left(đpcm\right)\)

Trước khi nhân thêm a+b+c thì phải chứng minh nó khác 0

\(\left(\frac{2+a}{2-a}-\frac{4a^2}{a^2-4}+\frac{a-2}{a+2}\right)\div\frac{a-3}{2a-a^2}\)(ĐK: \(a\ne0,a\ne\pm2,a\ne3\))

\(=\frac{\left(2+a\right)\left(2+a\right)+4a^2+\left(a-2\right)\left(2-a\right)}{\left(2-a\right)\left(2+a\right)}\div\frac{a-3}{a\left(2-a\right)}\)

\(=\frac{4+4a+a^2+4a^2-\left(a^2-4a+4\right)}{\left(2-a\right)\left(2+a\right)}.\frac{a\left(2-a\right)}{a-3}\)

\(=\frac{4a^2+8a}{\left(2-a\right)\left(2+a\right)}.\frac{a\left(2-a\right)}{a-3}\)

\(=\frac{4a^2}{a-3}\)

\(\left(\frac{2+a}{2-a}-\frac{4a^2}{a^2-4}+\frac{a-2}{a+2}\right)\div\frac{a-3}{2a-a^2}\)

\(=\left(\frac{2+a}{2-a}+\frac{4a^2}{4-a^2}+\frac{a-2}{2+a}\right)\div\frac{a-3}{2a-a^2}\)

\(=\left(\frac{\left(2+a\right)^2}{\left(2-a\right)\left(2+a\right)}+\frac{4a^2}{\left(2-a\right)\left(2+a\right)}+\frac{\left(a-2\right)\left(2-a\right)}{\left(2-a\right)\left(2+a\right)}\right)\div\frac{a-3}{2a-a^2}\)

\(=\left(\frac{4+4a+a^2+4a^2-a^2+4a-4}{\left(2-a\right)\left(2+a\right)}\right)\div\frac{a-3}{2a-a^2}\)

\(=\frac{4a^2+8a}{\left(2-a\right)\left(2+a\right)}\div\frac{a-3}{2a-a^2}\)

\(=\frac{4a\left(a+2\right)}{\left(2-a\right)\left(2+a\right)}\div\frac{a-3}{2a-a^2}\)

\(=\frac{4a}{2-a}.\frac{2a-a^2}{a-3}=\frac{4a\left(2a-a^2\right)}{\left(2-a\right)\left(a-3\right)}=\frac{4a^2\left(2-a\right)}{\left(2-a\right)\left(a-3\right)}=\frac{4a^2}{a-3}\)

\(x^2-\left(x+3\right)\left(3x+1\right)=\)\(9\)

\(\Leftrightarrow x^2-9-\left(x+3\right)\left(3x+1\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x+3\right)-\left(x+3\right)\left(3x+1\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(x-3-3x-1\right)=0\)

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

\(\Leftrightarrow\orbr{\begin{cases}x+3=0\\-2x-4=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-3\\-2x=4\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-3\\x=-2\end{cases}}}\)

Vậy phương trình có tập nghiệm \(S=\left\{-3;-2\right\}\)

\(x^3+4x+5=0\)

\(\Leftrightarrow\left(x^3+1\right)+\left(4x+4\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(x^2-x+1\right)+4\left(x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(x^2-x+1+4\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(x^2-x+5\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\x^2-x+5=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-1\\\left(x-\frac{1}{2}\right)^2+\frac{19}{4}=0\end{cases}}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=-1\\\left(x-\frac{1}{2}\right)^2=\frac{-19}{4}\left(vn\right)\end{cases}}\)(vn: vô nghiệm).\(\Leftrightarrow x=-1\)

Vậy phương trình có nghiệm duy nhất : \(x=-1\)

a) Xét tam giác \(MKN\)và tam giác \(MSP\): 

\(\widehat{M}\)chung

\(\widehat{MKN}=\widehat{MSP}\left(=90^o\right)\)

\(\Rightarrow\Delta MKN\)đồng dạng với \(\Delta MSP\)(g.g)

\(\Rightarrow\frac{MK}{MS}=\frac{MN}{MP}\)

\(\Rightarrow\frac{MK}{MN}=\frac{MS}{MP}\).

Xét tam giác \(MNP\)và tam giác \(MKS\):

\(\widehat{M}\)chung

\(\frac{MK}{MN}=\frac{MS}{MP}\)(cmt)

Suy ra tam giác \(MNP\)đồng dạng với tam giác \(MKS\)(c.g.c).

b), c) Tương tự.