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31 tháng 7 2018

BTVN nhiều nhỉ?

a,A=-1

b,B=2x-4y

c,C=2x^2-4

Bài 1: 

a: \(A=\left|2a-1\right|-2a\)

TH1: a>=1/2

A=2a-1-2a=-1

TH2: a<1/2

A=1-2a-2a=1-4a

b: \(B=x-2y-\left|x-2y\right|\)

TH1: x>=2y

A=x-2y-x+2y=0

TH2: x<2y

A=x-2y+x-2y=2x-4y

c: \(=x^2+\left|x^2-4\right|\)

TH1: x>=2 hoặc x<=-2

\(A=x^2+x^2-4=2x^2-4\)

TH2: -2<x<2

\(A=x^2+4-x^2=4\)

d: \(D=2x-1-\dfrac{\left|x-5\right|}{x-5}\)

TH1: x>5

\(D=2x-1-1=2x-2\)

TH2: x<5

D=2x-1+1=2x

17 tháng 8 2023

\(a,A=2\sqrt{20}-\dfrac{2}{\sqrt{3}+1}-\sqrt{80}+\sqrt{4+2\sqrt{3}}\\ =2.2\sqrt{5}-\dfrac{2\left(\sqrt{3}-1\right)}{\sqrt{3^2}-1}-4\sqrt{5}+\sqrt{\left(\sqrt{3}+1\right)^2}\\ =-\dfrac{2\left(\sqrt{3}-1\right)}{2}+\left|\sqrt{3}+1\right|\\ =-\sqrt{3}+1+\sqrt{3}+1\\ =2\)

\(B=\left(1+\dfrac{x+\sqrt{x}}{1+\sqrt{x}}\right)\left(1+\dfrac{x-\sqrt{x}}{1-\sqrt{x}}\right)\left(dk:x\ge0,x\ne1\right)\\ =\left(1+\dfrac{\sqrt{x}\left(1+\sqrt{x}\right)}{1+\sqrt{x}}\right)\left(1-\dfrac{\sqrt{x}\left(1-\sqrt{x}\right)}{1-\sqrt{x}}\right)\\ =\left(1+\sqrt{x}\right)\left(1-\sqrt{x}\right)\\ =1-x\)

\(b,A=4\sqrt{B}\Leftrightarrow4\sqrt{1-x}=2\\ \Leftrightarrow\sqrt{1-x}=\dfrac{1}{2}\\ \Leftrightarrow\left|1-x\right|=\dfrac{1}{4}\)

\(\Leftrightarrow1-x=\dfrac{1}{4}\\ \Leftrightarrow x=\dfrac{3}{4}\left(tm\right)\)

Vậy \(x=\dfrac{3}{4}\) thì \(A=4\sqrt{B}\).

17 tháng 8 2023

a) \(A=2\sqrt{20}-\dfrac{2}{\sqrt{3}+1}-\sqrt{80}+\sqrt{4+2\sqrt{3}}\)

\(A=2\cdot2\sqrt{5}-\dfrac{2\left(\sqrt{3}-1\right)}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}-4\sqrt{5}+\sqrt{\left(\sqrt{3}\right)^2+2\sqrt{3}\cdot1+1^2}\)

\(A=4\sqrt{5}-\dfrac{2\left(\sqrt{3}-1\right)}{2}-4\sqrt{5}+\sqrt{\left(\sqrt{3}+1\right)^2}\)

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

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

\(A=2\)

\(B=\left(1+\dfrac{x+\sqrt{x}}{1+\sqrt{x}}\right)\left(1+\dfrac{x-\sqrt{x}}{1-\sqrt{x}}\right)\)

\(B=\left[1+\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\right]\left[1-\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\right]\)

\(B=\left(1+\sqrt{x}\right)\left(1-\sqrt{x}\right)\)

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

\(B=1-x\)

b) Ta có: \(A=4\sqrt{B}\)

\(\Rightarrow2=4\sqrt{1-x}\)

\(\Leftrightarrow\sqrt{1-x}=\dfrac{1}{2}\)

\(\Leftrightarrow1-x=\dfrac{1}{4}\)

\(\Leftrightarrow x=1-\dfrac{1}{4}\)

\(\Leftrightarrow x=\dfrac{3}{4}\left(tm\right)\)

21 tháng 12 2023

Bài 3:
a) \(\sqrt{3x-2}=4\)
\(\sqrt{3x-2}=\sqrt{4^2}\)
\(3x-2=4^2=16\)
    \(3x=16+2=18\)
    \(x=18:3=6\)
    Vậy \(x=6\)
b)\(\sqrt{4x^2+4x+1}-11=5\)
\(\sqrt{\left(2x\right)^2+2\left(2x\right)\cdot1+1^2}-11=5\)
\(\sqrt{\left(2x+1\right)^2}-11=5\)
TH1:
\(\left(2x+1\right)-11=5\)
    \(2x+1=5+11=16\)
    \(2x=16-1=15\)
    \(x=15:2=7,5\)
TH2:
\(\left(2x+1\right)-11=-5\)
    \(2x-1=-5+11=6\)
    \(2x=6+1=7\)
    \(x=7:2=3,5\)
    Vậy \(x=\left\{7,5;3,5\right\}\) 
    (Câu này mình không chắc chắn lắm)   
    (Học sinh lớp 6 đang làm bài này)    

21 tháng 12 2023

Bài 4:

a: \(C=\left(\sqrt{x}-\dfrac{1}{\sqrt{x}}\right)\left(\dfrac{\sqrt{x}}{\sqrt{x}+1}+\dfrac{\sqrt{x}}{\sqrt{x}-1}\right)\)

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

\(=\dfrac{x-\sqrt{x}+x+\sqrt{x}}{\sqrt{x}}=\dfrac{2x}{\sqrt{x}}=2\sqrt{x}\)

b: C-6<0

=>C<6

=>\(2\sqrt{x}< 6\)

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

=>0<=x<9

Kết hợp ĐKXĐ, ta được: \(\left\{{}\begin{matrix}0< x< 9\\x\ne1\end{matrix}\right.\)

a: Ta có: \(\sqrt{4x+20}-3\sqrt{x+5}+\dfrac{4}{3}\sqrt{9x+45}=6\)

\(\Leftrightarrow2\sqrt{x+5}-3\sqrt{x+5}+4\sqrt{x+5}=6\)

\(\Leftrightarrow3\sqrt{x+5}=6\)

\(\Leftrightarrow x+5=4\)

hay x=-1

b: Ta có: \(\dfrac{1}{2}\sqrt{x-1}-\dfrac{3}{2}\sqrt{9x-9}+24\sqrt{\dfrac{x-1}{64}}=-17\)

\(\Leftrightarrow\dfrac{1}{2}\sqrt{x-1}-\dfrac{9}{2}\sqrt{x-1}+3\sqrt{x-1}=-17\)

\(\Leftrightarrow\sqrt{x-1}=17\)

\(\Leftrightarrow x-1=289\)

hay x=290

a) Ta có: \(\sqrt{25x+75}+3\sqrt{x-2}=2\sqrt{x-2}+\sqrt{9x-18}\)

\(\Leftrightarrow5\sqrt{x+3}+3\sqrt{x-2}=2\sqrt{x-2}+3\sqrt{x-2}\)

\(\Leftrightarrow\sqrt{25x+75}=\sqrt{4x-8}\)

\(\Leftrightarrow25x-4x=-8-75\)

\(\Leftrightarrow21x=-83\)

hay \(x=-\dfrac{83}{21}\)

b) Ta có: \(\sqrt{\left(2x-1\right)^2}=4\)

\(\Leftrightarrow\left|2x-1\right|=4\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-1=4\\2x-1=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=5\\2x=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)

c) Ta có: \(\sqrt{\left(2x+1\right)^2}=3x-5\)

\(\Leftrightarrow\left|2x+1\right|=3x-5\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+1=3x-5\left(x\ge-\dfrac{1}{2}\right)\\2x+1=5-3x\left(x< \dfrac{1}{2}\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-3x=-5-1\\2x+3x=5-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\left(nhận\right)\\x=\dfrac{4}{5}\left(loại\right)\end{matrix}\right.\)

d) Ta có: \(\sqrt{4x-12}-14\sqrt{\dfrac{x-2}{49}}=\sqrt{9x-18}+8\)

\(\Leftrightarrow2\sqrt{x-3}-2\sqrt{x-2}=3\sqrt{x-2}+8\)

\(\Leftrightarrow2\sqrt{x-3}-5\sqrt{x-2}=8\)

\(\Leftrightarrow4\left(x-3\right)+25\left(x-2\right)-20\sqrt{x^2-5x+6}=8\)

\(\Leftrightarrow4x-12+25x-50-8=20\sqrt{\left(x-2\right)\left(x-3\right)}\)

\(\Leftrightarrow20\sqrt{\left(x-2\right)\left(x-3\right)}=29x-70\)

\(\Leftrightarrow x^2-5x+6=\dfrac{\left(29x-70\right)^2}{400}\)

\(\Leftrightarrow x^2-5x+6=\dfrac{841}{400}x^2-\dfrac{203}{20}x+\dfrac{49}{4}\)

\(\Leftrightarrow\dfrac{-441}{400}x^2+\dfrac{103}{20}x-\dfrac{25}{4}=0\)

\(\Delta=\left(\dfrac{103}{20}\right)^2-4\cdot\dfrac{-441}{400}\cdot\dfrac{-25}{4}=-\dfrac{26}{25}\)(Vô lý)

vậy: Phương trình vô nghiệm

7 tháng 6 2017

a) \(\dfrac{\sqrt{16a^4b^6}}{\sqrt{128a^6b^6}}\)

\(=\dfrac{4a^2b^3}{8\sqrt{2}a^3b^3}\)

\(=\dfrac{1}{2\sqrt{2}a}\)

\(=\dfrac{\sqrt{2}}{4a}\)

b) \(\sqrt{\dfrac{x-2\sqrt{x}+1}{x+2\sqrt{x}+1}}\)

chịu đấy :v

c) \(\sqrt{\dfrac{\left(x-2\right)^2}{\left(3-x\right)^2}}+\dfrac{x^2-1}{x-3}\)

\(=\dfrac{x-2}{3-x}+\dfrac{x^2-1}{x-3}\)

\(=\dfrac{x-2}{-\left(x-3\right)}+\dfrac{x^2-1}{x-3}\)

\(=-\dfrac{x-2}{x-3}+\dfrac{x^2-1}{x-3}\)

\(=\dfrac{-\left(x-2\right)+x^2-1}{x-3}\)

\(=\dfrac{-x+1+x^2}{x-3}\)

d) \(\dfrac{x-1}{\sqrt{y}-1}\cdot\sqrt{\dfrac{\left(y-2\sqrt{y}+1^2\right)}{\left(x-1\right)^4}}\)

\(=\dfrac{x-1}{\sqrt{y}-1}\cdot\sqrt{\dfrac{y-2\sqrt{y}+1}{\left(x-1\right)^4}}\)

\(=\dfrac{x-1}{\sqrt{y}-1}\cdot\dfrac{\sqrt{y-2\sqrt{y}+1}}{\left(x-1\right)^2}\)

\(=\dfrac{1}{\sqrt{y}-1}\cdot\dfrac{\sqrt{y-2\sqrt{y}+1}}{x-1}\)

\(=\dfrac{\sqrt{y-2\sqrt{y}+1}}{\left(\sqrt{y}-1\right)\left(x-1\right)}\)

\(=\dfrac{\sqrt{y-2\sqrt{y}+1}}{x\sqrt{y}-\sqrt{y}-x+1}\)

e) \(4x-\sqrt{8}+\dfrac{\sqrt{x^3+2x^2}}{\sqrt{x+2}}\)

\(=4x-2\sqrt{2}+\dfrac{\sqrt{x^2\cdot\left(x+2\right)}}{\sqrt{x+2}}\)

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

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

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

8 tháng 6 2017

bạn ơi phần c mình sai đề bài.. bạn giúp mk giải lại đc k \(\sqrt{\dfrac{\left(x-2\right)^4}{\left(3-x\right)^2}}+\dfrac{x^2-1}{x-3}\)

a: \(=4a-4\sqrt{10a}-9\sqrt{10a}=4a-13\sqrt{10a}\)

b: \(=\sqrt{x}\left(4-\sqrt{2}\right)\cdot\sqrt{x}\left(1-\sqrt{2}\right)\)

\(=x\cdot\left(4-4\sqrt{2}-\sqrt{2}+2\right)\)

\(=\left(6-5\sqrt{2}\right)x\)

c: \(=\dfrac{2}{2x-1}\cdot x\sqrt{5}\cdot\left(2x-1\right)=2x\sqrt{5}\)