làm TT

Vì x = 4 là nghiệm của phương trình

nên thay x = 4 vào phương trình trên ta được :

\(-20-5=4a-9\)

\(\Leftrightarrow4a-9=-25\Leftrightarrow4a=-16\Leftrightarrow a=-4\)

Vậy a = -4

\(\left(2x-1\right)^2+2.\left(2x-1\right).\left(x+1\right)+\left(x+1\right)^2\)

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

\(=\left(3x\right)^2\)

\(=9x^2\)

\(4x^2-4x-5\left|2x-1\right|-5=0\)

\(\Leftrightarrow-5\left|2x-1\right|=5-4x^2+4x\)

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

\(\Leftrightarrow\left|2x-1\right|=\frac{4x\left(x-1\right)}{5}-1\)

TH1 : \(2x-1=\frac{4x\left(x-1\right)}{5}-1\Leftrightarrow2x=\frac{4x\left(x-1\right)}{5}\)

\(\Leftrightarrow10x=4x^2-4x\Leftrightarrow14x-4x^2=0\)

\(\Leftrightarrow-2x\left(2x-7\right)=0\Leftrightarrow x=0;x=\frac{7}{2}\)

TH2 : \(2x-1=-\left(\frac{4x\left(x-1\right)}{5}-1\right)\Leftrightarrow2x-1=-\frac{4x\left(x-2\right)}{5}+1\)

\(\Leftrightarrow2x-2=-\frac{4x\left(x-2\right)}{5}\Leftrightarrow10x-10=-4x^2+8x\)

\(\Leftrightarrow2x-10+4x^2=0\Leftrightarrow2\left(2x^2+x-5\ne0\right)=0\)tự chứng minh 

Vậy tập nghiệm của phương trình là S = { 0 ; 7/2 }

bằng2

1+1=2
#hỏi ngu thía!...:((
 

\(\frac{1}{x^2}+\frac{1}{9y^2}\ge2\sqrt{\frac{1}{x^2}.\frac{1}{9y^2}}=\frac{2}{3xy}=\frac{2}{3}\)

Dấu \(=\)xảy ra khi \(\hept{\begin{cases}\frac{1}{x^2}=\frac{1}{9y^2}\\xy=1\end{cases}}\Rightarrow\hept{\begin{cases}x=\sqrt{3}\\y=\frac{1}{\sqrt{3}}\end{cases}}\).

Ta có : \(7x^2+8xy+7y^2=10\)

\(\Rightarrow\left(x^2+2xy+y^2\right)+6\left(x^2+y^2\right)=10\)

\(\Rightarrow6\left(x^2+y^2\right)=10-\left(x+y\right)^2\)

\(\Rightarrow x^2+y^2=\frac{10-\left(x+y\right)^2}{6}=\frac{5}{3}-\frac{\left(x+y\right)^2}{6}\)

​Vì \(\left(x+y\right)^2\ge0\forall x,y\)\(\Rightarrow\frac{\left(x+y\right)^2}{6}\ge0\)

\(\Rightarrow x^2+y^2\le\frac{5}{3}\)

Dấu \("="\)xảy ra \(\Leftrightarrow\left(x+y\right)^2=0\)

\(\Leftrightarrow x+y=0\)

\(\Leftrightarrow x=-y\)

\(\Leftrightarrow7x^2-8x^2+7x^2=10\)

\(\Leftrightarrow6x^2=10\)

\(\Leftrightarrow x^2=\frac{5}{3}\)

\(\Leftrightarrow\)\(\hept{\begin{cases}x=\frac{5}{3}\\y=-\frac{5}{3}\end{cases}}\)

hoặc \(\hept{\begin{cases}x=-\frac{5}{3}\\y=\frac{5}{3}\end{cases}}\)

Ta dễ dàng chứng minh được : \(2xy\le x^2+y^2\forall x,y\)

\(\Rightarrow8xy\le4\left(x^2+y^2\right)\)

Ta có :\(7x^2+8xy+7y^2=7\left(x^2+y^2\right)+8xy=10\)

\(\Rightarrow7\left(x^2+y^2\right)=10-8xy\ge10-4\left(x^2+y^2\right)\)

\(\Rightarrow11\left(x^2+y^2\right)\ge10\)

\(\Rightarrow x^2+y^2\ge\frac{10}{11}\)

Dấu \("="\)xảy ra \(\Leftrightarrow x=y\)

\(\Leftrightarrow7x^2+8x^2+7x^2=10\)

\(\Leftrightarrow22x^2=10\)

\(\Leftrightarrow x^2=\frac{5}{11}\)

\(\Leftrightarrow\orbr{\begin{cases}x=y=\sqrt{\frac{5}{11}}\\x=y=-\sqrt{\frac{5}{11}}\end{cases}}\)

Vậy ...

\(3a^2-6ab+3b^2-12c^2\)

\(=3a^2-3ab-3ab+3b^2-12c^2\)

\(=\left(3a^2-3ab\right)-\left(3ab-3b^2\right)-12c^2\)

\(=3a\left(a-b\right)-3b\left(a-b\right)-12c^2\)

\(=\left(3a-3b\right)\left(a-b\right)-12c^2\)

\(=3\left(a-b\right)^2-12c^2\)

3a² - 6ab + 3b² - 12c²

= 3(a² - 2ab + b² - 4c²)

= 3[(a - b)² - (2c)²]

= 3(a - b + 2c)(a - b - 2c)