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16 tháng 12 2022

a: \(=3\sqrt{2x}-10\sqrt{2x}+21\sqrt{2x}=12\sqrt{2x}\)

b: \(=6-4\sqrt{3}+4\sqrt{3}-8=-2\)

c: \(=\sqrt{2}+1+2-\sqrt{2}=3\)

d: \(=\dfrac{1}{\sqrt{2}}\cdot\left(\sqrt{5}-\sqrt{3}-\sqrt{5}-\sqrt{3}+2\sqrt{3}\right)=0\)

f: \(=\sqrt{2}-8\sqrt{6}-\sqrt{2}+2\sqrt{6}=-6\sqrt{6}\)

a: Ta có: \(\sqrt{\left(5-\sqrt{19}\right)^2}-\sqrt{\left(4-\sqrt{19}\right)^2}\)

\(=5-\sqrt{19}-\sqrt{19}+4\)

\(=9-2\sqrt{19}\)

b: Ta có: \(\sqrt{\left(3-2\sqrt{2}\right)^2}-\sqrt{\left(2\sqrt{2}-3\right)^2}\)

\(=3-2\sqrt{2}-3+2\sqrt{2}\)

=0

 

 

AH
Akai Haruma
Giáo viên
2 tháng 10 2021

c.

Căn bậc 2 không xác định do $2-\sqrt{5}< 0$

d.

\(=\sqrt{(3+\sqrt{3})^2}(3+\sqrt{3})=|3+\sqrt{3}|(3+\sqrt{3})=(3+\sqrt{3})^2=12+6\sqrt{3}\)

e.

\(=(2-\sqrt{5})\sqrt{(2+\sqrt{5})^2}=(2-\sqrt{5})|2+\sqrt{5}|=(2-\sqrt{5})(2+\sqrt{5})=4-5=-1\)

AH
Akai Haruma
Giáo viên
3 tháng 8 2021

a. ĐKXĐ: $x\geq 1$

PT $\Leftrightarrow \frac{1}{2}\sqrt{x-1}-\frac{3}{2}.\sqrt{9}.\sqrt{x-1}+24.\sqrt{\frac{1}{64}}.\sqrt{x-1}=-17$

$\Leftrightarrow \frac{1}{2}\sqrt{x-1}-\frac{9}{2}\sqrt{x-1}+3\sqrt{x-1}=-17$

$\Leftrightarrow -\sqrt{x-1}=-17$

$\Leftrightarrow \sqrt{x-1}=17$

$\Leftrightarrow x-1=289$

$\Leftrightarrow x=290$

b. ĐKXĐ: $x\geq \frac{1}{2}$

PT $\Leftrightarrow \sqrt{9}.\sqrt{2x-1}-0,5\sqrt{2x-1}+\frac{1}{2}.\sqrt{25}.\sqrt{2x-1}+\sqrt{49}.\sqrt{2x-1}=24$

$\Leftrightarrow 3\sqrt{2x-1}-0,5\sqrt{2x-1}+2,5\sqrt{2x-1}+7\sqrt{2x-1}=24$
$\Leftrightarrow 12\sqrt{2x-1}=24$

$\Leftrihgtarrow \sqrt{2x-1}=2$

$\Leftrightarrow x=2,5$ (tm)

 

AH
Akai Haruma
Giáo viên
3 tháng 8 2021

c. ĐKXĐ: $x\geq 2$

PT $\Leftrightarrow \sqrt{36}.\sqrt{x-2}-15\sqrt{\frac{1}{25}}\sqrt{x-2}=4(5+\sqrt{x-2})$

$\Leftrightarrow 6\sqrt{x-2}-3\sqrt{x-2}=20+4\sqrt{x-2}$

$\Leftrightarrow \sqrt{x-2}=-20< 0$ (vô lý)

Vậy pt vô nghiệm

14 tháng 11 2018

Đề không khó, mỗi tội dài

14 tháng 11 2018

vậy thì bn làm hộ mik vs , mik cần gấp

AH
Akai Haruma
Giáo viên
30 tháng 9 2023

a.

$A=\sqrt{2-\sqrt{3}}+\sqrt{2+\sqrt{3}}$

$A\sqrt{2}=\sqrt{4-2\sqrt{3}}+\sqrt{4+2\sqrt{3}}$

$A\sqrt{2}=\sqrt{(\sqrt{3}-1)^2}+\sqrt{(\sqrt{3}+1)^2}$

$=|\sqrt{3}-1|+|\sqrt{3}+1|=\sqrt{3}-1+\sqrt{3}+1=2\sqrt{3}$

$\Rightarrow A=2\sqrt{3}: \sqrt{2}=\sqrt{6}$

---------------------

$B=\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}$

$B\sqrt{2}=\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}$

$B\sqrt{2}=\sqrt{(\sqrt{7}-1)^2}-\sqrt{(\sqrt{7}+1)^2}$

$=|\sqrt{7}-1|-|\sqrt{7}+1|=\sqrt{7}-1-(\sqrt{7}+1)=-2$

$\Rightarrow B=-2:\sqrt{2}=-\sqrt{2}$

30 tháng 9 2023

\(a,\sqrt{2-\sqrt{3}}+\sqrt{2+\sqrt{3}}\)

\(A-\sqrt{2}=\left(\sqrt{2-\sqrt{3}}-\sqrt{2+\sqrt{3}}\right)\cdot\sqrt{2}\\ =\sqrt{2-\sqrt{3}}\cdot\sqrt{2}-\sqrt{2+\sqrt{3}}\cdot\sqrt{2}\\ =\sqrt{\left(2-\sqrt{3}\right)\cdot2}-\sqrt{\left(2+\sqrt{3}\right)\cdot2}\\ =\sqrt{4-2\sqrt{3}}-\sqrt{4+2\sqrt{3}}\\ =\sqrt{3-2\sqrt{3}+1}-\sqrt{3+2\sqrt{3}+1}\\ =\sqrt{\left(\sqrt{3}-1\right)^2}-\sqrt{\left(\sqrt{3}+1\right)^2}\\ =\left|\sqrt{3}-1\right|-\left|\sqrt{3}+1\right|\\ =\sqrt{3}-1-\sqrt{3}-1\\ =-2\)

Ta có :

 \(A-\sqrt{2}=-2\\ \Leftrightarrow A=\dfrac{-2}{\sqrt{2}}=\dfrac{-\left(\sqrt{2}\right)^2}{\sqrt{2}}=-\sqrt{2}\)

__

C làm giống câu a, nhé.

__

\(\sqrt{\left(2\sqrt{5}+1\right)^2}-\sqrt{\left(\sqrt{5}-2\right)^2}\\ =\left|2\sqrt{5}+1\right|-\left|\sqrt{5}-2\right|\\ =2\sqrt{5}+1-\sqrt{5}+2\\ =3+\sqrt{5}\)

__

\(\sqrt{52-16\sqrt{3}}+\sqrt{\left(4\sqrt{3}-7\right)^2}\\ =\sqrt{48-2\cdot4\cdot\sqrt{3}\cdot2+4}+\left|4\sqrt{3}-7\right|\\ =\sqrt{\left(4\sqrt{3}\right)^2-2\cdot4\cdot\sqrt{3}\cdot2+2^2}+4\sqrt{3}-7\\ =\sqrt{\left(4\sqrt{3}-2\right)^2}+4\sqrt{3}-7\\ =4\sqrt{3}-2+4\sqrt{3}-7\\ =8\sqrt{3}-9\)

 

26 tháng 6 2021

\(A=2.\left|\left(-3\right)\right|^3+2.\left(-2\right)^2-4\left|\left(-2\right)^3\right|\)

\(=54+8-32=30\)

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

\(=5-2\sqrt{2}\)

\(C=\left|3-\sqrt{3}\right|-\left|1+\sqrt{3}\right|=3-\sqrt{3}-1-\sqrt{3}\)

\(=2-2\sqrt{3}\)

\(D=\left|5+\sqrt{6}\right|-\left|\sqrt{6}-5\right|=5+\sqrt{6}-5+\sqrt{6}\)

\(=2\sqrt{6}\)

\(E=\sqrt{15^2}-\sqrt{5^2}=15-5=10\)

26 tháng 6 2021

`A=2sqrt{(-3)^6}+2sqrt{(-2)^4}-4sqrt{(-2)^6}=2|(-3)^3|+2|(-2)^2|-4|(-2)^3|=54+8-32=30` $\\$ `B=sqrt{(sqrt2-2)^2}+sqrt{(sqrt2-3)^2}=2-sqrt2+3-sqrt2=5-2sqrt2` $\\$ `C=sqrt{(3-sqrt3)^2}-sqrt{(1+sqrt3)^2}=3-sqrt3-sqrt3-1=2-2sqrt3` $\\$ `D=sqrt{(5+sqrt6)^2}-sqrt{(sqrt6-sqrt5)^2}=5+sqrt6-5+sqrt6=2sqrt6` $\\$ `E=sqrt{17^2-8^2}-sqrt{3^2+4^2}=sqrt{289-64}-sqrt{9+16}=sqrt(225)-sqrt{25}=15-5=10`

a) Ta có: \(A=\dfrac{3+2\sqrt{3}}{\sqrt{3}}+\dfrac{2+\sqrt{2}}{\sqrt{2}+1}-\left(\sqrt{2}+\sqrt{3}\right)\)

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

=3

b) Ta có: \(B=\left(\dfrac{2}{\sqrt{3}-1}+\dfrac{3}{\sqrt{3}-2}+\dfrac{15}{3-\sqrt{3}}\right)\cdot\dfrac{1}{5+\sqrt{3}}\)

\(=\left[\sqrt{3}+1-3\left(2+\sqrt{3}\right)+\dfrac{15\left(3+\sqrt{3}\right)}{6}\right]\cdot\dfrac{1}{5+\sqrt{3}}\)

\(=\left(\sqrt{3}+1-6-3\sqrt{3}+\dfrac{5}{2}\left(3+\sqrt{3}\right)\right)\cdot\dfrac{1}{5+\sqrt{3}}\)

\(=\left(-5-2\sqrt{3}+\dfrac{15}{2}+\dfrac{5}{2}\sqrt{3}\right)\cdot\dfrac{1}{5+\sqrt{3}}\)

\(=\left(\dfrac{5}{2}+\dfrac{\sqrt{3}}{2}\right)\cdot\dfrac{1}{5+\sqrt{3}}=\dfrac{1}{2}\)

 

a) Ta có: \(\left(7\sqrt{48}+3\sqrt{27}-2\sqrt{12}\right)\cdot\sqrt{3}\)

\(=\left(7\cdot4\sqrt{3}+3\cdot3\sqrt{3}-2\cdot2\sqrt{3}\right)\cdot\sqrt{3}\)

\(=33\sqrt{3}\cdot\sqrt{3}\)

=99

b) Ta có: \(\left(12\sqrt{50}-8\sqrt{200}+7\sqrt{450}\right):\sqrt{10}\)

\(=\left(12\cdot5\sqrt{2}-8\cdot10\sqrt{2}+7\cdot15\sqrt{2}\right):\sqrt{10}\)

\(=\dfrac{85\sqrt{2}}{\sqrt{10}}=\dfrac{85}{\sqrt{5}}=17\sqrt{5}\)

c) Ta có: \(\left(2\sqrt{6}-4\sqrt{3}+5\sqrt{2}-\dfrac{1}{4}\sqrt{8}\right)\cdot3\sqrt{6}\)

\(=\left(2\sqrt{6}-4\sqrt{3}+5\sqrt{2}-\dfrac{1}{4}\cdot2\sqrt{2}\right)\cdot3\sqrt{6}\)

\(=\left(2\sqrt{6}-4\sqrt{3}+3\sqrt{2}\right)\cdot3\sqrt{6}\)

\(=36-36\sqrt{2}+18\sqrt{3}\)

d) Ta có: \(3\sqrt{15\sqrt{50}}+5\sqrt{24\sqrt{8}}-4\sqrt{12\sqrt{32}}\)

\(=3\cdot\sqrt{75\sqrt{2}}+5\cdot\sqrt{48\sqrt{2}}-4\sqrt{48\sqrt{2}}\)

\(=3\cdot5\sqrt{2}\cdot\sqrt{\sqrt{2}}+4\sqrt{3}\sqrt{\sqrt{2}}\)

\(=15\sqrt{\sqrt{8}}+4\sqrt{\sqrt{18}}\)

2 tháng 7 2021

a,=\(\left(28\sqrt{3}+9\sqrt{3}-4\sqrt{3}\right).\sqrt{3}\)

   \(=28.3+9.3-4.3=99\)

b,\(=\left(60\sqrt{2}-80\sqrt{2}+175\sqrt{2}\right):\sqrt{10}\)

  \(=155\sqrt{2}:\sqrt{10}=\dfrac{155}{\sqrt{5}}\)

Bài 2: Thực hiện phép tínha) \(\sqrt{5}-\sqrt{48}+5\sqrt{27}-\sqrt{45}\)b) \(\left(\sqrt{5}+\sqrt{2}\right)\left(3\sqrt{2}-1\right)\)c) \(3\sqrt{50}-2\sqrt{75}-4\frac{\sqrt{54}}{\sqrt{3}}-3\sqrt{\frac{1}{3}}\)d) \(\sqrt{\left(\sqrt{3}-3\right)^2}+\sqrt{4-2\sqrt{3}}\)e) \(\sqrt{48-2\sqrt{135}}-\sqrt{45}+\sqrt{18}\)f) \(\frac{5\sqrt{2}-2\sqrt{5}}{\sqrt{5}-\sqrt{2}}+\frac{6}{2-\sqrt{10}}-\frac{20}{\sqrt{10}}\)Bài 3: Thực hiện phép...
Đọc tiếp

Bài 2: Thực hiện phép tính

a) \(\sqrt{5}-\sqrt{48}+5\sqrt{27}-\sqrt{45}\)

b) \(\left(\sqrt{5}+\sqrt{2}\right)\left(3\sqrt{2}-1\right)\)

c) \(3\sqrt{50}-2\sqrt{75}-4\frac{\sqrt{54}}{\sqrt{3}}-3\sqrt{\frac{1}{3}}\)

d) \(\sqrt{\left(\sqrt{3}-3\right)^2}+\sqrt{4-2\sqrt{3}}\)

e) \(\sqrt{48-2\sqrt{135}}-\sqrt{45}+\sqrt{18}\)

f) \(\frac{5\sqrt{2}-2\sqrt{5}}{\sqrt{5}-\sqrt{2}}+\frac{6}{2-\sqrt{10}}-\frac{20}{\sqrt{10}}\)

Bài 3: Thực hiện phép tính

a) \(\sqrt{9-4\sqrt{5}}\)

b) \(2\sqrt{3}+\sqrt{48}-\sqrt{75}-\sqrt{243}\)

c) \(\sqrt{4+\sqrt{8}}.\sqrt{2+\sqrt{2+\sqrt{2}}}.\sqrt{2-\sqrt{2+\sqrt{2}}}\)

d) \(\sqrt{3+2\sqrt{2}}-\sqrt{6-4\sqrt{2}}\)

e) \(\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}+\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}-\frac{\sqrt{5}+1}{\sqrt{5}-1}\)

f*) \(\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}\)

Bài 4: Rút gọn

a) \(3\sqrt{2x}-5\sqrt{8x}+7\sqrt{18x}\)

b) \(\left(2\sqrt{3}+\sqrt{4}\right)\left(\sqrt{3}-2\right)\)

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

d) \(\sqrt{4-\sqrt{15}}-\sqrt{4+\sqrt{15}}+\sqrt{6}\)

e) \(\left(\frac{5-\sqrt{5}}{\sqrt{5}}-2\right)\left(\frac{4}{1+\sqrt{5}}+4\right)\)

f) \(\frac{1}{5}\sqrt{50}-2\sqrt{96}-\frac{\sqrt{30}}{\sqrt{15}}+12\sqrt{\frac{1}{6}}\)

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