\(\frac{\sqrt{7-4\sqrt{3}}}{\sqrt{6}-2\sqrt{2}}\)

\(=\frac{\sqrt{4-2.2\sqrt{3}+3}}{\sqrt{2}\left(\sqrt{3}-2\right)}\)

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

\(=\frac{\left|2-\sqrt{3}\right|}{-\sqrt{2}\left(2-\sqrt{3}\right)}\)

\(=\frac{2-\sqrt{3}}{-\sqrt{2}\left(2-\sqrt{3}\right)}\)

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

\(\frac{\sqrt{7-4\sqrt{3}}}{\sqrt{6}-2\sqrt{2}}\)

\(=\frac{\sqrt{4-2.2\sqrt{3}+3}}{\sqrt{2}\left(\sqrt{3}-2\right)}\)

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

\(=\frac{\left|2-\sqrt{3}\right|}{-\sqrt{2}\left(2-\sqrt{3}\right)}\)

\(=\frac{2-\sqrt{3}}{-\sqrt{2}\left(2-\sqrt{3}\right)}\)

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

Biểu thức có nghĩa khi:

\(\hept{\begin{cases}x-1\ge0\\x-3\ge0\end{cases}}\)

\(\Leftrightarrow\)\(\hept{\begin{cases}x\ge1\\x\ge3\end{cases}}\)

\(\Leftrightarrow\)\(x\ge3\)

dinh công hải

khó,ko giải đc

a) ktra lại đề

b)  \(3x^2+6xy+3y^2-3z^2=3\left(x^2+2xy+y^2-z^2\right)=3\left[\left(x+y\right)^2-z^2\right]\)

\(=3\left(x+y+z\right)\left(x+y-z\right)\)

c)  \(x^2-2xy+y^2-z^2+2zt-t^2=\left(x-y\right)^2-\left(z-t\right)^2=\left(x-y-z+t\right)\left(x-y+z-t\right)\)

d)  \(2x^2+4x-2-2y^2=2\left(x^2-y^2+2x-1\right)\)

e)  \(2xy-x^2-y^2+16=16-\left(x-y\right)^2=\left(4-x+y\right)\left(4+x-y\right)\)

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

g)  \(x^4+4=x^4+4x^2+4-4x^2=\left(x^2+2\right)-4x^2=\left(x^2-2x+2\right)\left(x^2+2x+2\right)\)

h)  \(x^3+2x^2+2x+1=\left(x+1\right)\left(x^2-x+1\right)+2x\left(x+1\right)=\left(x+1\right)\left(x^2+x+1\right)\)

câu đề là a)     x² + 4x – y² + 4  nha

\(\frac{1}{2}\sqrt{6}>\frac{1}{6}\sqrt{2}\)

\(\frac{1}{2}\sqrt{6}>\frac{1}{6}\sqrt{2}\)

A B C D H

Áp dụng hệ thức lượng ta có:

    AD² = AH.AC

=>  AC = AD²/AH = 10

Áp dụng Pytago ta có:

   AD² + DC² = AC²

=>  DC² = AC² - AD² = 64

=> DC = 8

=> AB = DC = 8

\(B=\sqrt[3]{5+2\sqrt{13}}+\sqrt[3]{5-2\sqrt{13}}\)

Áp dụng \(\left(a+b\right)^3=a^3+b^3+3ab\left(a+b\right)\)ta có:

\(B^3=5+2\sqrt{13}+5-2\sqrt{13}+3B\sqrt[3]{25-52}\)

\(=10-9B\)

Giải PT: \(B^3+9B-10=0\Leftrightarrow B^3-1+9B-9=0\)\(\Leftrightarrow\left(B-1\right)\left(B^2+2B+1\right)+9\left(B-1\right)=0\)

\(\Leftrightarrow\left(B-1\right)\left(B^2+2B+10\right)=0\)\(\Leftrightarrow\orbr{\begin{cases}B-1=0\\B^2+2B+1+9=0\end{cases}\Leftrightarrow\orbr{\begin{cases}B=1\\\left(B+1\right)^2=-9\left(L\right)\end{cases}}}\)

Vậy \(B=1\)

À chết mình làm nhầm, phải là \(\left(B-1\right)\left(B^2+B+1\right)\) nha, \(\left(B-1\right)\left(B^2+B+2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}B=1\\B^2+2.\frac{1}{2}B+\frac{1}{4}-\frac{1}{4}+2=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}B=1\\\left(B+\frac{1}{2}\right)^2+\frac{7}{4}=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}B=1\\\left(B+\frac{1}{2}\right)^2=-\frac{7}{4}\left(L\right)\end{cases}}\)

\(\sqrt{3+\sqrt{x}}=3\)   \(;\)  \(x\ge0\)

\(\Leftrightarrow\) \(3+\sqrt{x}=9\)

\(\Leftrightarrow\) \(\sqrt{x}=6\)

\(\Leftrightarrow\)   \(x=36\left(tm\text{đ}k\right)\)

Vậy chọn đáp án C

\(\sqrt{3+\sqrt{x}}=3\)   \(;\)  \(x\ge0\)

\(\Leftrightarrow\) \(3+\sqrt{x}=9\)

\(\Leftrightarrow\) \(\sqrt{x}=6\)

\(\Leftrightarrow\)   \(x=36\left(tm\text{đ}k\right)\)

Vậy chọn đáp án C

\(2\sqrt[3]{3}>\sqrt[3]{23}\)

So sánh \(2\sqrt[3]{3}\) và  \(\sqrt[3]{23}\)

Ta có :

\(2\sqrt[3]{3}=\sqrt[3]{2^3}.\sqrt[3]{3}=\sqrt[3]{24}\)

Vì \(\sqrt[3]{24}>\sqrt[3]{23}\)

Nên \(2\sqrt[3]{3}>\sqrt[3]{23}\)