căn 15 < căn 16=4

căn 8 < căn 9 bằng 3 

mà 4=3=7 suy ra 7>căn 15 cộng căn 8

\(A=\sqrt{49}.\sqrt{144}+\sqrt{256}:\sqrt{64}\)

\(A=7.12+\sqrt{4}\)

\(A=84+2\)

\(A=86\)

\(B=72:\sqrt{2^2.36.3^2}-\sqrt{225}\)

\(B=72:\sqrt{36.36}-15\)

\(B=2-15=-13\)

a) đk: \(x\ge2\)

\(\sqrt{x^2-2x+1}+\sqrt{x^2-4x+4}=3\)

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

\(\Leftrightarrow\left|x-1\right|+\left|x-2\right|=3\)

\(\Leftrightarrow\hept{\begin{cases}\left(x-1\right)+\left(x-2\right)=3\\\left(x-1\right)-\left(x-2\right)=3\end{cases}\Leftrightarrow\hept{\begin{cases}x-1+x-2=3\\x-1-x+2=3\end{cases}}}\)\(\Leftrightarrow\hept{\begin{cases}2x-3=3\\0x+1=3\end{cases}\Leftrightarrow\hept{\begin{cases}2x=6\\0x=2\end{cases}\Leftrightarrow}x=3}\)(tm)

Vậy ....

b) \(\sqrt{2x-2+2\sqrt{2x-3}}+\sqrt{2x+13+8\sqrt{2x-3}}=5\)

Đặt \(\sqrt{2x-3}=t\left(t\ge0\right)\)

\(\Rightarrow\sqrt{t^2-1+2t}+\sqrt{t^2+16+8t}=5\)

\(\Leftrightarrow\sqrt{\left(t-1\right)^2}+\sqrt{\left(t+4\right)^2}=5\)

\(\Leftrightarrow t-1+t+4=5\)

\(\Leftrightarrow2t=2\Leftrightarrow t=1\left(tm\right)\)

\(\Rightarrow\sqrt{2x-3}=1\Leftrightarrow2x-3=1^2\Leftrightarrow x=2\)

Vậy...

\(x^3=\left(\sqrt{5}+\sqrt{3}\right)^3=\sqrt{5^3}+3.5.\sqrt{3}+3.\sqrt{5}.3+\sqrt{3^3}\)

\(=14\sqrt{5}+18\sqrt{3}\)

\(x^2=\left(\sqrt{5}+\sqrt{3}\right)^2=8+2\sqrt{15}\)

\(\frac{1}{x^2}=\frac{1}{8+2\sqrt{15}}=\frac{8-2\sqrt{15}}{8^2-4.15}=\frac{4-\sqrt{15}}{2}\)

\(\Leftrightarrow\frac{4}{x^2}=8-2\sqrt{15}\)

\(\Leftrightarrow x^2+\frac{4}{x^2}=16\)

\(\Leftrightarrow x^4-16x^2+4=0\)

Ta có đpcm. 

\(f,\sqrt[3]{5\sqrt{2}+7}-\sqrt[3]{5\sqrt{2}-7}\)

\(=\sqrt[3]{3\sqrt{2}+2\sqrt{2}+1+6}-\sqrt[3]{3\sqrt{2}+2\sqrt{2}-1-6}\)

\(=\sqrt[3]{6+2\sqrt{2}+1+3\sqrt{2}}-\sqrt[3]{-6+2\sqrt{2}-1+3\sqrt{2}}\)

\(=\sqrt[3]{\left(1+\sqrt{2}\right)^3}-\sqrt[3]{-\left(1-\sqrt{2}\right)^3}\)

\(=1+\sqrt{2}+1-\sqrt{2}=2\)

\(g,\sqrt[3]{20+14\sqrt{2}}+\sqrt[3]{20-14\sqrt{2}}\)

\(=\sqrt[3]{\left(2+\sqrt{2}\right)^3}+\sqrt[3]{\left(2-\sqrt{2}\right)^3}\)

\(=2+\sqrt{2}+2-\sqrt{2}=4\)

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

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

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

 \(A\cdot\sqrt{x}=9\Leftrightarrow\frac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}}\cdot\sqrt{x}-9=0\)

\(\Leftrightarrow\left(\sqrt{x}-2\right)\left(\sqrt{x}+4\right)=0\Leftrightarrow x=4\left(tm\right)\)

Vậy ... 

\(\frac{AB}{AC}=\frac{3}{4}\Leftrightarrow AB=\frac{3}{4}AC\)

\(BC^2=AB^2+AC^2=\left(\frac{3}{4}AC\right)^2+AC^2=\frac{25}{16}AC^2\)

\(\Leftrightarrow AC=\frac{4}{5}BC\)

Suy ra \(AC=100\left(cm\right),AB=75\left(cm\right)\)

\(HB=\frac{AB^2}{BC}=\frac{75^2}{125}=45\left(cm\right),HC=BC-HB=80\left(cm\right)\)

Từ \(\frac{AB}{AC}=\frac{3}{4}\Rightarrow AB=\frac{3}{4}AC\)

Xét tam giác ABC vuông tại A có: \(BC^2=AB^2+AC^2\)(Pi - ta - go)

=> \(AB^2+AC^2=125^2=15625\)

<=> \(\left(\frac{3}{4}AC\right)^2+AC^2=15625\)

<=> \(\frac{25}{16}AC^2=15625\)=> \(AC=100\)(cm)  => \(AB=75\)(cm)

Xét tam giác ABC vuông tại A có AH là đường cao

=> \(AB^2=BH.BC\) (hệ thức lượng) =>> \(BH=\frac{AB^2}{BC}=\frac{75^2}{125}=45\)(cm)

\(AC^2=HC.BC\) => \(HC=\frac{AC^2}{BC}=\frac{100^2}{125}=80\)(cm)

a) \(B=\frac{1}{\frac{1}{4}\sqrt{\frac{1}{4}}+27}=\frac{1}{\frac{1}{4}\cdot\frac{1}{2}+27}=\frac{1}{\frac{1}{8}+27}=\frac{1}{\frac{217}{8}}=\frac{8}{217}\)

b) \(A=\frac{x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}-\frac{1}{2-\sqrt{x}}-\frac{\sqrt{x}}{\sqrt{x}+3}\)

\(A=\frac{x-9+\sqrt{x}+3-\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)

\(A=\frac{x-6+\sqrt{x}-x+2\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\)

\(A=\frac{3\sqrt{x}-6}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}=\frac{3\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}=\frac{3}{\sqrt{x}+3}\)

c)) Với x \(\ge\)0 và x \(\ne\)4 (1)

Ta có: \(A>\frac{1}{2}\) <=> \(\frac{3}{\sqrt{x}+3}>\frac{1}{2}\)

<=> \(\sqrt{x}+3< 6\) <=> \(\sqrt{x}< 3\) <=> \(x< 9\) (2)

Từ (1) và (2) => \(0\le x< 9\)và x khác 4

d) Ta có : \(C=B:A=\frac{1}{x\sqrt{x}+27}:\frac{3}{\sqrt{x}+3}\)

\(C=\frac{1}{\left(\sqrt{x}+3\right)\left(x-3\sqrt{x}+9\right)}\cdot\frac{\sqrt{x}+3}{3}\)

\(C=\frac{1}{3\left(x-3\sqrt{x}+9\right)}=\frac{1}{3\left(x-3\sqrt{x}+\frac{9}{4}\right)+\frac{81}{4}}=\frac{1}{\left(\sqrt{x}-\frac{3}{2}\right)^2+\frac{81}{4}}\)

Do \(\left(\sqrt{x}-\frac{3}{2}\right)^2+\frac{81}{4}\ge\frac{81}{4}\) => \(C\le\frac{1}{\frac{81}{4}}=\frac{4}{81}\)

Dấu "=" xảy ra<=> \(\sqrt{x}-\frac{3}{4}=0\) <=> \(x=\frac{9}{16}\)

Vậy MaxC = 4/81 <=> x = 9/16

ok k nha

a) Xét (O) có 

$\widehat{PFE}$ là góc nội tiếp chắn nửa đường tròn

nên $\widehat{PFE}={90}^0$(Hệ quả góc nội tiếp)

hay $\widehat{PFN}={90}^0$

Xét tứ giác PFMN có 

$\widehat{PFN}$ và $\widehat{PMN}$ là hai góc cùng nhìn cạnh PN

$\widehat{PFN}=\widehat{PMN}\left(={90}^0\right)$

Do đó: PFMN là tứ giác nội tiếp(Dấu hiệu nhận biết tứ giác nội tiếp)