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

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

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

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

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

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

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

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

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

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

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

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

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

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

Ta có: \(x=\frac{1}{\sqrt[3]{3-2\sqrt{2}}}+\sqrt[3]{3-2\sqrt{2}}\)

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

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

=> \(x^3=\frac{3+2\sqrt{2}}{9-8}+3-2\sqrt{2}+3x\)

=> \(x^3=3+2\sqrt{2}+3-2\sqrt{2}+3x\)

=> \(x^3=3x+6\)

Do đó: \(P=\left(2x^3-6x+2008\right)^{2021}\)

\(P=\left[2\left(3x+6\right)-6x+2008\right]^{2021}\)

\(P=\left(6x+12-6x+2008\right)^{2021}=2020^{2021}\)

Xét tam giác \(AHC\)vuông tại \(H\)đường cao \(HN\):

\(AH^2=AN.AC\).

Xét tam giác \(AHB\)vuông tại \(H\)đường cao \(HM\):

\(AH^2=AM.AB\)

Suy ra \(AN.AC=AM.AB\Leftrightarrow\frac{AN}{AB}=\frac{AM}{AC}\)

Suy ra \(\Delta ANM~\Delta ABC\left(c.g.c\right)\)

\(\Rightarrow\widehat{ANM}=\widehat{ABC}\)

suy ra \(\widehat{ABC}+\widehat{MNC}=180^o\)

suy ra \(BMNC\)là tứ giác nội tiếp..

ối giồi giải mấy bài trẻ con này thì chắc ko bao giờ vươn ra thế giới, há há