$20\sqrt{\dfrac{x-2}{x 1}}-5\sqrt{\dfrac{x 2}{x-1}}=-28\sqrt[4]{\dfrac{x^2-4}{x^2-1}}$

LL

Ling ling 2k7

14 tháng 10 2021

Giúp với

1) Thu gọn A

$A=\left(\sqrt{x}-\dfrac{x+2}{\sqrt{x}+1}\right):\left(\dfrac{\sqrt{x}}{\sqrt{x}+1}-\dfrac{\sqrt{x}-4}{1-x}\right)$

2) Tính A biết $x=\left(\dfrac{2-\sqrt{5}}{2+\sqrt{5}}-\dfrac{2+\sqrt{5}}{2-\sqrt{5}}\right):\sqrt{20}$

1

NH

Nguyễn Hoàng Minh

14 tháng 10 2021

$A=\dfrac{x+\sqrt{x}-x-2}{\sqrt{x}+1}:\dfrac{x-\sqrt{x}+\sqrt{x}-4}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\\ A=\dfrac{\sqrt{x}-2}{\sqrt{x}+1}\cdot\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{\sqrt{x}-1}{\sqrt{x}+2}$

$x=\dfrac{9-4\sqrt{5}-9-4\sqrt{5}}{\left(2+\sqrt{5}\right)\left(2-\sqrt{5}\right)}:2\sqrt{5}=\dfrac{-8\sqrt{5}}{-2\sqrt{5}}=4\\ \Leftrightarrow\sqrt{x}=2\\ \Leftrightarrow A=\dfrac{2-1}{2+2}=\dfrac{1}{4}$

H

hoàng

15 tháng 9 2023

4.

$A=\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}\right)-\left(\dfrac{1}{x+\sqrt{x}}\right).\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{2}{x-1}\right)$

a. Rút gọn A.

b. Tính x khi $A=\dfrac{1}{2}$

5. CMR

$\left(\dfrac{\sqrt{30}}{\sqrt{3}}-\dfrac{\sqrt{20}}{\sqrt{2}}-\dfrac{6}{\sqrt{6}}\right).\sqrt{4+\sqrt{15}}=2$

nhanh lên nha

2

H

hoàng

15 tháng 9 2023

help

NL

Nguyễn Lê Phước Thịnh

15 tháng 9 2023

loading... => đề sai rồi bạn

KA

Khánh An Ngô

22 tháng 7 2023

a) $\sqrt{4x^2-9}=2\sqrt{x+3}$

b) $\sqrt{4x+20}+3\sqrt{\dfrac{x-5}{9}}-\dfrac{1}{3}\sqrt{9x-45}=4$

c) $\dfrac{2}{3}\sqrt{9x-9}-\dfrac{1}{4}\sqrt{16x-16}+27\sqrt{\dfrac{x-1}{81}}=4$

d)$5\sqrt{\dfrac{9x-27}{25}}-7\sqrt{\dfrac{4x-12}{9}}-7\sqrt{x^2-9}+18\sqrt{\dfrac{9x^2-81}{81}}=0$

2

VV

Võ Việt Hoàng

22 tháng 7 2023

$a) \sqrt{4x^2− 9} = 2\sqrt{x + 3}$

$ĐK:x\ge\dfrac{3}{2}$

$pt\Leftrightarrow4x^2-9=4\left(x+3\right)$

$\Leftrightarrow4x^2-9=4x+12$

$\Leftrightarrow4x^2-4x-21=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1-\sqrt{22}}{2}\left(l\right)\\x=\dfrac{1+\sqrt{22}}{2}\left(tm\right)\end{matrix}\right.$

$b)\sqrt{4x-20}+3.\sqrt{\dfrac{x-5}{9}}-\dfrac{1}{3}\sqrt{9x-45}=4$

$ĐK:x\ge5$

$pt\Leftrightarrow2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4$

$\Leftrightarrow2\sqrt{x-5}=4\Leftrightarrow\sqrt{x-5}=2$

$\Leftrightarrow x-5=4\Leftrightarrow x=9\left(tm\right)$

VV

Võ Việt Hoàng

22 tháng 7 2023

$c)\dfrac{2}{3}\sqrt{9x-9}-\dfrac{1}{4}\sqrt{16x-16}+27.\sqrt{\dfrac{x-1}{81}}=4$

ĐK:x>=1

$pt\Leftrightarrow2\sqrt{x-1}-\sqrt{x-1}+3\sqrt{x-1}=4$

$\Leftrightarrow4\sqrt{x-1}=4\Leftrightarrow\sqrt{x-1}=1$

$\Leftrightarrow x-1=1\Leftrightarrow x=2\left(tm\right)$

$d)5\sqrt{\dfrac{9x-27}{25}}-7\sqrt{\dfrac{4x-12}{9}}-7\sqrt{x^2-9}+18\sqrt{\dfrac{9x^2-81}{81}}=0$

$ĐK:x\ge3$

$pt\Leftrightarrow3\sqrt{x-3}-\dfrac{14}{3}\sqrt{x-3}-7\sqrt{x^2-9}+6\sqrt{x^2-9}=0$

$\Leftrightarrow-\dfrac{5}{3}\sqrt{x-3}-\sqrt{x^2-9}=0\Leftrightarrow\dfrac{5}{3}\sqrt{x-3}+\sqrt{x^2-9}=0$

$\Leftrightarrow(\dfrac{5}{3}+\sqrt{x+3})\sqrt{x-3}=0$

$\Leftrightarrow\sqrt{x-3}=0$ (vì $\dfrac{5}{3}+\sqrt{x+3}>0$)

$\Leftrightarrow x-3=0\Leftrightarrow x=3\left(nhận\right)$

BB

Biện Bạch Ngọc

15 tháng 9 2017

1/ Thu gọn các biểu thức:

A=$\sqrt{\dfrac{3\sqrt{3}-4}{2\sqrt{3}+1}}-\sqrt{\dfrac{\sqrt{3}+4}{5-2\sqrt{3}}}$

B=$\dfrac{x\sqrt{x}-2x+28}{x-3\sqrt{x}-4}-\dfrac{\sqrt{x}-4}{\sqrt{x}+1}+\dfrac{\sqrt{x}+8}{4-\sqrt{x}}$

1

NL

Nguyễn Lê Phước Thịnh

27 tháng 5 2022

a: $A=\sqrt{\dfrac{3\sqrt{3}-4}{2\sqrt{3}+1}}-\sqrt{\dfrac{4+\sqrt{3}}{5-2\sqrt{3}}}$

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

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

$=\dfrac{\sqrt{3}-1-\sqrt{3}-1}{\sqrt{2}}=-\sqrt{2}$

b: $B=\dfrac{x\sqrt{x}-2x+28}{\left(\sqrt{x}-4\right)\left(\sqrt{x}+1\right)}-\dfrac{x-16}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-4\right)}-\dfrac{\left(\sqrt{x}+8\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-4\right)\left(\sqrt{x}+1\right)}$

$=\dfrac{x\sqrt{x}-2x+28-x+16-x-9\sqrt{x}-8}{\left(\sqrt{x}-4\right)\left(\sqrt{x}+1\right)}$

$=\dfrac{x\sqrt{x}-4\sqrt{x}-9\sqrt{x}+36}{\left(\sqrt{x}-4\right)\left(\sqrt{x}+1\right)}=\dfrac{x-9}{\sqrt{x}+1}$

a : $\dfrac{3}{\sqrt{x}-5}+\dfrac{20-2\sqrt{x}}{x-25}$với x ≥ 0 x ≠ 25b : $\dfrac{\sqrt{x}}{\sqrt{x}-3}+\dfrac{2\sqrt{x}-2}{x-9}$với x ≥ 0 x ≠ 9c : $\dfrac{\sqrt{x}-1}{\sqrt{x}+2}+\dfrac{5\sqrt{x}-2}{x-4}$với x ≥ 0 x ≠ 4d : $\left(\dfrac{x-2}{x+2\sqrt{x}}+\dfrac{1}{\sqrt{x}+2}\right).\dfrac{\sqrt{x}+1}{\sqrt{x}-1}$với ≥ 0 x ≠...

M

manh

7 tháng 10 2023

Đọc tiếp

5

PN

Phương Nhi

7 tháng 10 2023

$a,\dfrac{3}{\sqrt{x}-5}+\dfrac{20-2\sqrt{x}}{x-25}\\ =\dfrac{3}{\sqrt{x}-5}+\dfrac{20-2\sqrt{x}}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\\ =\dfrac{3\left(\sqrt{x}+5\right)}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}+\dfrac{20-2\sqrt{x}}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\\ =\dfrac{3\sqrt{x}+15+20-2\sqrt{x}}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\\ =\dfrac{\sqrt{x}+35}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}$

PN

Phương Nhi

7 tháng 10 2023

$b,\dfrac{\sqrt{x}}{\sqrt{x}-3}+\dfrac{2\sqrt{x}-2}{x-9}\\ =\dfrac{\sqrt{x}}{\sqrt{x}-3}+\dfrac{2\sqrt{x}-2}{x-9}\\ =\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{x-9}+\dfrac{2\sqrt{x}-2}{x-9}\\ =\dfrac{x+3\sqrt{x}+2\sqrt{x}-2}{x-9}\\ =\dfrac{x-5\sqrt{x}-2}{x-9}$

M

manh

15 tháng 10 2023

a $\sqrt{4x-20}+\sqrt{x-5}-\dfrac{1}{3}\sqrt{9x-45}=4$

b $\sqrt{x-1}+\sqrt{4x-4}-\sqrt{25x-25}=4$

c $\dfrac{1}{3}\sqrt{x-2}-\dfrac{2}{3}\sqrt{9x-18}+6\sqrt{\dfrac{x-2}{81}=-4}$

d $\sqrt{9x+27}+4\sqrt{x+3}-\dfrac{3}{4}\sqrt{16x+48}=0$

3

NL

Nguyễn Lê Phước Thịnh

15 tháng 10 2023

a: ĐKXĐ: x-5>=0

=>x>=5

$\sqrt{4x-20}+\sqrt{x-5}-\dfrac{1}{3}\cdot\sqrt{9x-45}=4$

=>$2\sqrt{x-5}+\sqrt{x-5}-\dfrac{1}{3}\cdot3\sqrt{x-5}=4$

=>$2\sqrt{x-5}=4$

=>x-5=4

=>x=9(nhận)

b: ĐKXĐ: x-1>=0

=>x>=1

$\sqrt{x-1}+\sqrt{4x-4}-\sqrt{25x-25}=4$

=>$\sqrt{x-1}+2\sqrt{x-1}-5\sqrt{x-1}=4$

=>$-2\sqrt{x-1}=4$

=>$\sqrt{x-1}=-2$(vô lý)

Vậy: Phương trình vô nghiệm

c: ĐKXĐ: x-2>=0

=>x>=2

$\dfrac{1}{3}\sqrt{x-2}-\dfrac{2}{3}\cdot\sqrt{9x-18}+6\cdot\sqrt{\dfrac{x-2}{81}}=-4$

=>$\dfrac{1}{3}\sqrt{x-2}-\dfrac{2}{3}\cdot3\sqrt{x-2}+6\cdot\dfrac{\sqrt{x-2}}{9}=-4$

=>$\sqrt{x-2}\left(\dfrac{1}{3}-2+\dfrac{2}{3}\right)=-4$

=>$-\sqrt{x-2}=-4$

=>x-2=16

=>x=18(nhận)

d: ĐKXĐ: x+3>=0

=>x>=-3

$\sqrt{9x+27}+4\sqrt{x+3}-\dfrac{3}{4}\cdot\sqrt{16x+48}=0$

=>$3\sqrt{x+3}+4\sqrt{x+3}-\dfrac{3}{4}\cdot4\sqrt{x+3}=0$

=>$4\sqrt{x+3}=0$

=>x+3=0

=>x=-3(nhận)

NV

Nhật Văn

15 tháng 10 2023

a) $\sqrt{4x-20}+\sqrt{x-5}-\dfrac{1}{3}\sqrt{9x-45}=4$

= $2\sqrt{x-5}+\sqrt{x-5}-\dfrac{1}{3}\sqrt{9\left(x-5\right)}=4$

= $2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4$

= $2\sqrt{x-5}=4$

= $\sqrt{x-5}=2$

= $\left|x-5\right|=4$

=> $x-5=\pm4$

$x=\pm4+5$

$x=9;x=1$

Vậy x=9; x=1

CP

Cần Phải Biết Tên

5 tháng 9 2018

Thu gọn biểu thức

A=$\sqrt{\dfrac{3\sqrt{3}}{2\sqrt{3}+1}}-\sqrt{\dfrac{\sqrt{3}+4}{5-2\sqrt{3}}}$

B=$\dfrac{x\sqrt{x}-2x+28}{x-3\sqrt{x}-4}-\dfrac{\sqrt{x}-4}{\sqrt{x}+1}+\dfrac{\sqrt{x}+8}{4-\sqrt{x}}\left(x\ge0,x\ne16\right)$

1

NL

Nguyễn Lê Phước Thịnh

31 tháng 8 2022

$A=\sqrt{\dfrac{18-3\sqrt{3}}{11}}-\sqrt{2+\sqrt{3}}$

$=\dfrac{\sqrt{11\left(18-3\sqrt{3}\right)}}{11}-\dfrac{\sqrt{4+2\sqrt{3}}}{\sqrt{2}}$

$=\dfrac{\sqrt{11\left(18-3\sqrt{3}\right)}}{11}-\dfrac{\sqrt{3}+1}{\sqrt{2}}$

$=\dfrac{\sqrt{11\left(18-3\sqrt{3}\right)}}{11}-\dfrac{\sqrt{6}+\sqrt{2}}{2}$

$=\dfrac{2\sqrt{11\left(18-3\sqrt{3}\right)}-11\sqrt{6}-11\sqrt{2}}{22}$

b: $=\dfrac{x\sqrt{x}-2x+28-x+16-x-9\sqrt{x}-8}{\left(\sqrt{x}-4\right)\left(\sqrt{x}+1\right)}$

$=\dfrac{x\sqrt{x}-4x-9\sqrt{x}+36}{\left(\sqrt{x}-4\right)\left(\sqrt{x}+1\right)}=\dfrac{x-9}{\sqrt{x}+1}$

KT

Kim Tuyến

1 tháng 10 2021

giải các phương trình sau:

a. $2\sqrt{2x}-5\sqrt{8x}+7\sqrt{18x}=28$

b. $\sqrt{4x-20}+\sqrt{x-5}-\dfrac{1}{3}\sqrt{9x-45}=4$

c. $\sqrt{\dfrac{3x-2}{x+1}}=3$

1

AH

Akai Haruma

Giáo viên

1 tháng 10 2021

Lời giải:

a. ĐKXĐ: $x\geq 0$

$2\sqrt{2x}-5\sqrt{8x}+7\sqrt{18x}=28$

$\Leftrightarrow 2\sqrt{2x}-10\sqrt{2x}+21\sqrt{2x}=28$

$\Leftrightarrow 13\sqrt{2x}=28$

$\Leftrightarrow \sqrt{2x}=\frac{28}{13}$

$\Leftrightarrow 2x=\frac{784}{169}$

$\Leftrightarrow x=\frac{392}{169}$

b. ĐKXĐ: $x\geq 5$

PT $\Leftrightarrow \sqrt{4}.\sqrt{x-5}+\sqrt{x-5}-\frac{1}{3}.\sqrt{9}.\sqrt{x-5}=4$

$\Leftrightarrow 2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4$

$\Leftrightarrow 2\sqrt{x-5}=4$

$\Leftrightarrow \sqrt{x-5}=2$

$\Leftrightarrow x-5=4$

$\Leftrightarrow x=9$ (tm)

c. ĐKXĐ: $x\geq \frac{2}{3}$ hoặc $x< -1$

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

$\Rightarrow 3x-2=9(x+1)$

$\Leftrightarrow x=\frac{-11}{6}$ (tm)

DT

Đặng Tuyết Đoan

5 tháng 8 2021

Bài 1: Tìm x, biết

a)$2\sqrt{9x-27}-\dfrac{1}{5}\sqrt{25x-75}-\dfrac{1}{7}\sqrt{49x-147}=20$

b) $\sqrt{9x+18}-5\sqrt{x+2}+\dfrac{4}{5}\sqrt{25x+50}=6$

c)$\sqrt{16x-16}-\sqrt{9x-9}+\sqrt{4x-4}+\sqrt{x-1}=8$

d) $\sqrt{x+2\sqrt{x-1}}-\sqrt{x-2\sqrt{x-1}}=2$

1

NL

Nguyễn Lê Phước Thịnh

5 tháng 8 2021

a) Ta có: $2\sqrt{9x-27}-\dfrac{1}{5}\sqrt{25x-75}-\dfrac{1}{7}\sqrt{49x-147}=20$

$\Leftrightarrow6\sqrt{x-3}-\sqrt{x-3}-\sqrt{x-3}=20$

$\Leftrightarrow4\sqrt{x-3}=20$

$\Leftrightarrow x-3=25$

hay x=28

b) Ta có: $\sqrt{9x+18}-5\sqrt{x+2}+\dfrac{4}{5}\sqrt{25x+50}=6$

$\Leftrightarrow3\sqrt{x+2}-5\sqrt{x+2}+4\sqrt{x+2}=6$

$\Leftrightarrow2\sqrt{x+2}=6$

$\Leftrightarrow x+2=9$

hay x=7