S=A-B+C

\(=\frac{\sqrt7-2}{\sqrt7\left(\sqrt7+2\right)}-\frac{1}{\sqrt7+2}+\frac{1}{\sqrt7}\)

\(=\frac{\sqrt7-2-\sqrt7+\sqrt7+2}{\sqrt7\left(\sqrt7+2\right)}=\frac{\sqrt7}{\sqrt7\left(\sqrt7+2\right)}=\frac{1}{\sqrt7+2}\)

\(=\frac{\sqrt7-2}{3}\)

Bài 1:

a)  \(B=\sqrt{1-4x+4x^2}\)

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

         \(=\left|1-2x\right|\)

Nếu  \(x\le\frac{1}{2}\)thì:  \(B=1-2x\)

Nếu  \(x>\frac{1}{2}\)thì:  \(B=2x-1\)

b)  Tại \(x=-7\)thì:  \(B=1-2.\left(-7\right)=15\)

Bài 2:

\(\sqrt{7+4\sqrt{3}}+\sqrt{7-4\sqrt{3}}\)

\(=\sqrt{\left(\sqrt{3}\right)^2+2.\sqrt{3}.2+2^2}+\sqrt{2^2-2.2.\sqrt{3}+\left(\sqrt{3}\right)^2}\)

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

\(=\sqrt{3}+2+2-\sqrt{3}=4\) (đpcm)

có bộ gõ kí hiệu Toán mà :))

ĐK : a >= 0 ; a khác 36

\(K=\left[\frac{a+14\sqrt{a}+100}{\left(\sqrt{a}-6\right)\left(\sqrt{a}+7\right)}+\frac{\left(\sqrt{a}+6\right)\left(\sqrt{a}-6\right)}{\left(\sqrt{a}-6\right)\left(\sqrt{a}+7\right)}-\frac{\left(\sqrt{a}-7\right)\left(\sqrt{a}+7\right)}{\left(\sqrt{a}-6\right)\left(\sqrt{a}+7\right)}\right]\div\left(\frac{\sqrt{a}-6}{\sqrt{a}-6}-\frac{\sqrt{a}-7}{\sqrt{a}-6}\right)\)

\(=\frac{a+14\sqrt{a}+100+a-36-a+49}{\left(\sqrt{a}-6\right)\left(\sqrt{a}+7\right)}\div\frac{1}{\sqrt{a}-6}\)

\(=\frac{a+14\sqrt{a}+113}{\left(\sqrt{a}-6\right)\left(\sqrt{a}+7\right)}\cdot\left(\sqrt{a}-6\right)=\frac{a+14\sqrt{a}+113}{\sqrt{a}+7}\)

Để K = 2 thì \(\frac{a+14\sqrt{a}+113}{\sqrt{a}+7}=2\Rightarrow a+14\sqrt{a}+113=2\sqrt{a}+14\Leftrightarrow a+12\sqrt{a}+99=0\)

Với a >= 0 thì \(a+12\sqrt{a}+99\ge99>0\)=> Không có giá trị x thỏa mãn K = 2

Ta có : \(K=\frac{a+14\sqrt{a}+113}{\sqrt{a}+7}=\frac{\left(a+14\sqrt{a}+49\right)+64}{\sqrt{a}+7}=\frac{\left(\sqrt{a}+7\right)^2+64}{\sqrt{a}+7}\)

\(=\left(\sqrt{a}+7\right)+\frac{64}{\sqrt{a}+7}\ge2\sqrt{\left(\sqrt{a}+7\right)\cdot\frac{64}{\sqrt{a}+7}}=16\)( bđt AM-GM )

Dấu "=" xảy ra <=> \(\sqrt{a}+7=\frac{64}{\sqrt{a}+7}\Rightarrow a=1\left(tm\right)\). Vậy MinK = 16

bấm máy tính là ra liền đó chị ơi

a, 2

b, \(\frac{1}{2}\)

c,\(\sqrt{3}\)

mk k chắc lém nhưng bn cho mk nha mk tl đầu tiên

Ỏ

Bạn tốt qá he

\(x,y \geq 0\)

a: \(=9\sqrt{2}-4\sqrt{2}+4\sqrt{2}+9\sqrt{2}=18\sqrt{2}\)

b: \(=8\sqrt{3}-12\sqrt{3}+5\sqrt{3}+2\sqrt{3}=3\sqrt{3}\)

c: \(=2\sqrt{21}\)

Bài 2:

a: ĐKXĐ: x>=0

\(\sqrt{3x}-5\sqrt{12x}+7\cdot\sqrt{27x}=12\)

=>\(\sqrt{3x}-5\cdot2\sqrt{3x}+7\cdot3\sqrt{3x}=12\)

=>\(12\sqrt{3x}=12\)

=>\(\sqrt{3x}=1\)

=>3x=1

=>x=1/3(nhận)

Bài 1:

a: \(A=\left(\sqrt{\frac23}+\sqrt{\frac{50}{3}}-\sqrt{24}\right)\cdot\sqrt6\)

\(=\left(\frac{2\sqrt6}{6}+\sqrt{\frac{100}{6}}-2\sqrt6\right)\cdot\sqrt6\)

\(=2+\sqrt{100}-2\cdot6=2+10-12=0\)

b: \(B=\left(\frac{\sqrt{14}-\sqrt7}{\sqrt2-1}+\frac{\sqrt{15}-\sqrt5}{\sqrt3-1}\right):\frac{1}{\sqrt7-\sqrt5}\)

\(=\left(\frac{\sqrt7\left(\sqrt2-1\right)}{\sqrt2-1}+\frac{\sqrt5\left(\sqrt3-1\right)}{\sqrt3-1}\right)\cdot\left(\sqrt7-\sqrt5\right)\)

\(=\left(\sqrt7+\sqrt5\right)\left(\sqrt7-\sqrt5\right)\)

=7-5

=2