\(=2,35693368\)

cách làm kìa ai chẳng biết kết quả

a)\(\sqrt{\left(x+3\right)\left(x+2\right)}+\sqrt{\left(x+3\right)\left(x-1\right)}=2\sqrt{\left(x+3\right)^2}\)

\(\Leftrightarrow\sqrt{\left(x+3\right)\left(x+2\right)}+\sqrt{\left(x+3\right)\left(x-1\right)}-2\sqrt{\left(x+3\right)^2}=0\)

\(\Leftrightarrow\sqrt{x+3}\left(\sqrt{x+2}+\sqrt{x-1}-2\sqrt{x+3}\right)=0\)

\(\Rightarrow\orbr{\begin{cases}\sqrt{x+3}=0\\\sqrt{x+2}+\sqrt{x-1}=2\sqrt{x+3}\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x+3=0\\2x+1+2\sqrt{\left(x-1\right)\left(x+2\right)}=4\left(x+3\right)\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=-3\\2\sqrt{\left(x-1\right)\left(x+2\right)}=2x+11\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=-3\\4\left(x-1\right)\left(x+2\right)=4x^2+44x+121\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=-3\\-40x=129\end{cases}}\Rightarrow x=-3\) (thỏa)

b)\(\frac{3x}{\sqrt{3x+10}}=\sqrt{3x+1}-1\)

Đk:\(x\ge-\frac{1}{3}\)

\(pt\Leftrightarrow\frac{3x}{\sqrt{3x+10}}+1=\sqrt{3x+1}\)

\(\Leftrightarrow\frac{3x}{\sqrt{3x+10}}+1-\left(\frac{3}{5}x+1\right)=\sqrt{3x+1}-\left(\frac{3}{5}x+1\right)\)

\(\Leftrightarrow\frac{3x}{\sqrt{3x+10}}-\frac{3}{5}x=\frac{3x+1-\left(\frac{3}{5}x+1\right)^2}{\sqrt{3x+1}+\frac{3}{5}x+1}\)

\(\Leftrightarrow\frac{3x\left(5-\sqrt{3x+10}\right)}{5\sqrt{3x+10}}=\frac{-\frac{9}{25}x\left(x-5\right)}{\sqrt{3x+1}+\frac{3}{5}x+1}\)

\(\Leftrightarrow\frac{3x\cdot\frac{25-3x-10}{5+\sqrt{3x+10}}}{5\sqrt{3x+10}}-\frac{-\frac{9}{25}x\left(x-5\right)}{\sqrt{3x+1}+\frac{3}{5}x+1}=0\)

\(\Leftrightarrow\frac{3x\cdot\frac{-3\left(x-5\right)}{5+\sqrt{3x+10}}}{5\sqrt{3x+10}}-\frac{-\frac{9}{25}x\left(x-5\right)}{\sqrt{3x+1}+\frac{3}{5}x+1}=0\)

\(\Leftrightarrow x\left(x-5\right)\left(\frac{\frac{-9}{5+\sqrt{3x+10}}}{5\sqrt{3x+10}}-\frac{-\frac{9}{25}}{\sqrt{3x+1}+\frac{3}{5}x+1}\right)=0\)

Dễ thấy: \(\frac{\frac{-9}{5+\sqrt{3x+10}}}{5\sqrt{3x+10}}-\frac{-\frac{9}{25}}{\sqrt{3x+1}+\frac{3}{5}x+1}< 0\)

\(\Rightarrow\orbr{\begin{cases}x=0\\x-5=0\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x=0\\x=5\end{cases}}\)

Đk: \(\orbr{\begin{cases}x\le0\\x\ge4\end{cases}}\)
( đặt nhân tử chung ta có x=0, nghiệm thứ 2 \(\sqrt{x}\)là -1/56 nên loại)
Vậy x=0

\(2x^4-13x^3+24x^2-13x+2=0\)

\(\Leftrightarrow2x^4-8x^3+2x^2-5x^3+20x^2-5x+2x^2-8x+2=0\)

\(\Leftrightarrow2x^2\left(x^2-4x+1\right)-5x\left(x^2-4x+1\right)+2\left(x^2-4x+1\right)=0\)

\(\Leftrightarrow2x^2\left(x^2-4x+1\right)-5x\left(x^2-4x+1\right)+2\left(x^2-4x+1\right)=0\)

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

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

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

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

\(\Rightarrow\orbr{\begin{cases}x=2;x=\frac{1}{2}\\x=\frac{4\pm\sqrt{12}}{2}\end{cases}}\)

bạn có thể giair theo cacsh đối xứng đươcj ko cái mà chia cả 2 vế cho x² rồi đặt ý làm phuền bạn

a) \(\frac{3+2\sqrt{2}}{1+\sqrt{2}}=\frac{\left(1+\sqrt{2}\right)^2}{1+\sqrt{2}}=1+\sqrt{2}\)

b)\(\frac{4\sqrt{3}+2}{2\sqrt{3}+1}=\frac{2.\left(2\sqrt{3}+1\right)}{2\sqrt{3}+1}=2\)

c)\(\sqrt{300}-3\sqrt{10}+\sqrt{40}=10\sqrt{3}-3\sqrt{10}+2\sqrt{10}=10\sqrt{3}-\sqrt{10}\)

... dúng thì ủng hộ nha ...
Kết bạn với mình .. ;) ;)

a, \(\frac{3+2\sqrt{2}}{1+\sqrt{2}}=\frac{5,828427125}{2,4142133562}\)

b, \(\frac{4\sqrt{3}+2}{2\sqrt{3}+1}=\frac{8,92820323}{4,464101615}\)

c, \(\sqrt{300}-3\sqrt{10}+\sqrt{40}=14,15823042\)

P/s; Ko chắc đâu nhé. Sai thì bỏ qua cho mình nhé, mình mới lớp 5 lên lớp 6 thôi

2). X-1= (√x-1).(√x+1)

3) a+√a=  √a (√a+1)

Cac bn nho ung ho mk nha

Áp dụng BĐT Cauchy-Schwarz ta có:

\(VT^2=\left(\sqrt{2010-x}+\sqrt{x-2008}\right)^2\)

\(\le\left(1+1\right)\left(2010-x+x-2008\right)\)

\(=2\cdot\left(2010-2008\right)=2\cdot2=4\)

\(\Rightarrow VT^2\le4\Rightarrow VT\le2\)

Lại có: \(VP=x^2-4018x+4036083\)

\(=x^2-4018x+4036081+2\)

\(=\left(x-2009\right)^2+2\ge2\)

Suy ra \(VT\le VP=2\) xảy ra khi \(VT=VP=2\)

\(\Rightarrow\left(x-2009\right)^2+2=2\Rightarrow x-2009=0\Rightarrow x=2009\)