ĐK : x ≥ 0 , x ≠ 1

\(=\left[\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+1}-\frac{x+2}{\sqrt{x}+1}\right]\left[\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\frac{\sqrt{x}-4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right]\)

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

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

\(\frac{a}{b}=\frac{1}{4}\Rightarrow b=4a\)

\(a,b>0\)

\(B=\frac{-\sqrt{a}-\sqrt{b}}{a-\sqrt{ab}}=\frac{-\sqrt{a}-\sqrt{4a}}{a-\sqrt{a.4a}}=\frac{-\sqrt{a}-2\sqrt{a}}{a-2a}=\frac{-3\sqrt{a}}{-a}=\frac{3}{\sqrt{a}}=1\)

\(\Rightarrow a=9\Rightarrow b=36\).

e chịu hoy cj

Bài này bạn tự vẽ hình nha,

Gợi ý: b là cạnh đối diện góc B nên AC=b

          c là cạnh đối diện góc C nên AB=c

Do tam giác ABC vuông tại A nên

\(sin\widehat{B}=\frac{AC}{BC}=\frac{b}{BC}\)

\(\Rightarrow\)\(\frac{b}{sin\widehat{B}}=\frac{b}{\frac{b}{BC}}=b.\frac{BC}{b}=BC\)  ( 1 )

\(sin\widehat{C}=\frac{AB}{BC}=\frac{c}{BC}\)

\(\Rightarrow\)\(\frac{c}{sin\widehat{C}}=\frac{c}{\frac{c}{BC}}=c.\frac{BC}{c}=BC\) ( 2 )

Từ ( 1 ) và ( 2 )

\(\Rightarrow\)\(\frac{b}{sin\widehat{B}}=\frac{c}{sin\widehat{C}}\)

Bạn nhớ kỹ cái này nhé:

sin : Đi - học ( Đối - huyền)

cos : Không - hư ( Kề - huyền )

tan : Đoàn - kết ( Đối - kề )

cot : Kết - đoàn ( Kề - đối )

a) ta có 2=1+1=\(\sqrt{1}\)+1

Vì \(\sqrt{1}\)<\(\sqrt{2}\)nên \(\sqrt{1}\)+1  <   \(\sqrt{2}\)+1

 \(\Rightarrow\)2<\(\sqrt{2}\)+1

b) ta có: 12=\(\sqrt{25}\)+\(\sqrt{49}\)

\(\Rightarrow\)\(\hept{\begin{cases}\sqrt{24}< \sqrt{25}\\\sqrt{45}< \sqrt{49}\end{cases}}\Rightarrow\)\(\sqrt{24}+\sqrt{45}< \sqrt{25}+\sqrt{49}\)hay \(\sqrt{24}+\sqrt{45}\)<12

chịu thôi

\(x^4+2x^3+2x^2-2x+1=\left(x^3+x\right)\sqrt{\frac{1-x^2}{x}}\)ĐK: \(0< x< 1\)

\(\Leftrightarrow\left(x^4+2x^3+x^2\right)+\left(x^2-2x+1\right)=x\left(x^2+1\right)\sqrt{\frac{\left(1-x\right)\left(1+x\right)}{x}}\)

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

Đặt  \(\hept{\begin{cases}x\left(x+1\right)=a>0\\1-x=b>0\end{cases}}\)\(\Rightarrow x^2+1=a+b\)

\(\Rightarrow\left(1\right)\Leftrightarrow a^2+b^2=\left(a+b\right)\sqrt{ab}\)

\(\Leftrightarrow a^2+b^2-a\sqrt{ab}-b\sqrt{ab}=0\)

\(\Leftrightarrow a\sqrt{a}\left(\sqrt{a}-\sqrt{b}\right)+b\sqrt{b}\left(\sqrt{b}-\sqrt{a}\right)=0\)

\(\Leftrightarrow\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a^3}-\sqrt{b^3}\right)=0\)

\(\Leftrightarrow\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)=0\)

\(\Leftrightarrow\left(\sqrt{a}-\sqrt{b}\right)^2\left(a+\sqrt{ab}+b\right)=0\)

Vì \(a+\sqrt{ab}+b=\left(\sqrt{a}+\frac{\sqrt{b}}{2}\right)^2+\frac{3\sqrt{b}}{4}>0;\forall a,b>0\)

\(\Rightarrow\left(\sqrt{a}-\sqrt{b}\right)^2=0\)

\(\Leftrightarrow a=b\)

\(\Rightarrow x^2+x=1-x\)

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

\(\Delta=8\)

=> pt có 2 nghiệm pb \(^{\orbr{\begin{cases}x=-1+\sqrt{2}\left(tm\right)\\x=-1-\sqrt{2}\left(loai\right)\end{cases}}}\)

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