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\(ĐKXĐ:\hept{\begin{cases}a>0\\a\ne1\end{cases}}\)

\(P=\frac{2a+4}{a\sqrt{a}-1}+\frac{\sqrt{a}+2}{a+\sqrt{a}+1}-\frac{2}{\sqrt{a}-1}\)

\(=\frac{2a+4+\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)-2\left(a+\sqrt{a}+1\right)}{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}\)

\(=\frac{2a+4+a+\sqrt{a}-2-2a-2\sqrt{a}-2}{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}\)

\(=\frac{a-\sqrt{a}}{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}\)

\(=\frac{\sqrt{a}}{a+\sqrt{a}+1}\)

Ta có:

\(P=\frac{2a+4}{a\sqrt{a}-1}+\frac{\sqrt{a}+2}{a+\sqrt{a}+1}-\frac{2}{\sqrt{a}-1}\)

\(P=\frac{2a+4+\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)-2\left(a+\sqrt{a}+1\right)}{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}\)

\(P=\frac{2a+4+a+\sqrt{a}-2-2a-2\sqrt{a}-2}{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}\)

\(P=\frac{a-\sqrt{a}}{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}\)

\(P=\frac{\sqrt{a}}{a+\sqrt{a}+1}\)

\(\frac{a\sqrt{a}+b\sqrt{b}}{\sqrt{a}+\sqrt{b}}=\frac{\left(\sqrt{a}+\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)}{\sqrt{a}+\sqrt{b}}=a+\sqrt{ab}+b\)

Thiếu đkxđ: \(a,b>0\)

Ta có: \(\hept{\begin{cases}x^2+xy+y=1\\x+xy+y^2=1\end{cases}}\)

\(\Leftrightarrow x^2+xy+y+x+xy+y^2=2\)

\(\Leftrightarrow\left(x^2+2xy+y^2\right)+\left(x+y\right)=2\)

\(\Leftrightarrow\left(x+y\right)^2+\left(x+y\right)=2\)

\(\Leftrightarrow\left(x+y\right)\left(x+y+1\right)=2\)

Sau đó xét các TH

\(hpt< =>\hept{\begin{cases}\left(x+y\right)^2+\left(x+y\right)=2\\x^2-y^2-\left(x-y\right)=0\end{cases}}< =>\hept{\begin{cases}\left(x+y\right)^2+\left(x+y\right)=2\\\left(x+y\right)\left(x-y\right)-\left(x-y\right)=0\end{cases}}\)

Đặt \(\left\{x+y;x-y\right\}\rightarrow\left\{a;b\right\}\)Suy ra \(\hept{\begin{cases}a^2+a-2=0\\ab-b=0\end{cases}< =>\hept{\begin{cases}a^2+a-2=0\\b\left(a-1\right)=0\end{cases}< =>\hept{\begin{cases}b=0\\a=1\end{cases}}}}\)

\(< =>\hept{\begin{cases}x+y=1\\x-y=0\end{cases}< =>\hept{\begin{cases}x=1-y\\1-y-y=0\end{cases}< =>\hept{\begin{cases}x=1-y\\y=\frac{1}{2}\end{cases}}< =>x=y=\frac{1}{2}}}\)

Bài 9: 

a) đk: \(x\ge0\)

Ta có: \(3+\sqrt{x}\ne5\)

\(\Leftrightarrow\sqrt{x}\ne2\)

\(\Rightarrow x\ne4\)

Vậy \(\hept{\begin{cases}x\ge0\\x\ne4\end{cases}}\)

b) \(\sqrt{x^2-6x+9}=3\)

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

\(\Leftrightarrow\left|x-3\right|=3\)

\(\Leftrightarrow\orbr{\begin{cases}x-3=3\\x-3=-3\end{cases}}\Rightarrow\orbr{\begin{cases}x=6\\x=0\end{cases}}\)

Bài 9.

a) \(3+\sqrt{x}=5\)

ĐK : x ≥ 0

<=> \(\sqrt{x}=2\)

<=> \(x=4\)( tm )

Vậy x = 4

b) \(\sqrt{x^2-6x+9}=3\)

<=> \(\sqrt{\left(x-3\right)^2}=3\)

<=> \(\left|x-3\right|=3\)

<=> \(\orbr{\begin{cases}x-3=3\\x-3=-3\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=6\\x=0\end{cases}}\)

Bài 10.

\(A=\sqrt{x^2-2x+5}\)

=> A² = x² - 2x + 5

          = ( x² - 2x + 1 ) + 4

          = ( x - 1 )² + 4 ≥ 4 ∀ x

Dấu "=" xảy ra khi x = 1

=> A² ≥ 4

=> A ≥ 2

=> MinA = 2 <=> x = 1

b) \(B=\sqrt{\frac{x^2}{4}-\frac{x}{6}+1}\)

=> B² = \(\frac{1}{4}x^2-\frac{1}{6}x+1\)

          = \(\left(\frac{1}{4}x^2-\frac{1}{6}x+\frac{1}{36}\right)+\frac{35}{36}\)

          = \(\left(\frac{1}{2}x-\frac{1}{6}\right)^2+\frac{35}{36}\ge\frac{35}{36}\forall x\)

Dấu "=" xảy ra khi x = 1/3

=> B² ≥ 35/36

=> B ≥ \(\frac{\sqrt{35}}{6}\)

=> MinB = \(\frac{\sqrt{35}}{6}\)<=> x = 1/3

a) \(A=5x-\sqrt{4x^2-4x+1}\)

\(=5x-\sqrt{\left(2x-1\right)^2}\)

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

+) Với x < 1/2

A = 5x - [ -( 2x - 1 ) ] = 5x - ( 1 - 2x ) = 5x - 1 + 2x = 7x - 1

+) Với x ≥ 1/2

A = 5x - ( 2x - 1 ) = 5x - 2x + 1 = 3x + 1

b) Với x = -2 < 1/2

=> A = 7.(-2) - 1 = -14 - 1 = -15

a) \(\sqrt{\left(5-\sqrt{3}\right)^2}=\left|5-\sqrt{3}\right|=5-\sqrt{3}\)

b) \(\sqrt{\left(1-\sqrt{2}\right)^2}=\left|1-\sqrt{2}\right|=-\left(1-\sqrt{2}\right)=\sqrt{2}-1\)( vì 1 < √2 )

c) \(\sqrt{\left(\sqrt{3}-2\right)^2}=\left|\sqrt{3}-2\right|=-\left(\sqrt{3}-2\right)=2-\sqrt{3}\)( vì √3 < 2 )