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7 tháng 10 2018

\(a.\sqrt{\left(2x-5\right)^2}=7\Leftrightarrow2x-5=7\)

\(\Leftrightarrow2x=12\Leftrightarrow x=6\)

b, Máy mình lỗi font nên không làm đc

7 tháng 10 2018

bẠN VIẾT RA GIẤY RỒI CHỤP GIÚP MÌNH ĐC KH A

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

17 tháng 12 2023

a: ĐKXĐ: \(x\in R\)

\(\sqrt{\left(2x+3\right)^2}=5\)

=>|2x+3|=5

=>\(\left[{}\begin{matrix}2x+3=5\\2x+3=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=2\\2x=-8\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=1\left(nhận\right)\\x=-4\left(nhận\right)\end{matrix}\right.\)

b: ĐKXĐ: \(x\in R\)

\(\sqrt{9\left(x-2\right)^2}=18\)

=>\(\sqrt{9}\cdot\sqrt{\left(x-2\right)^2}=18\)

=>\(3\cdot\left|x-2\right|=18\)

=>\(\left|x-2\right|=6\)

=>\(\left[{}\begin{matrix}x-2=6\\x-2=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\left(nhận\right)\\x=-4\left(nhận\right)\end{matrix}\right.\)
c: ĐKXĐ: x>=2

\(\sqrt{9x-18}-\sqrt{4x-8}+3\sqrt{x-2}=40\)

=>\(3\sqrt{x-2}-2\sqrt{x-2}+3\sqrt{x-2}=40\)

=>\(4\sqrt{x-2}=40\)

=>\(\sqrt{x-2}=10\)

=>x-2=100

=>x=102(nhận)

d: ĐKXĐ: \(x\in R\)

\(\sqrt{4\left(x-3\right)^2}=8\)

=>\(\sqrt{\left(2x-6\right)^2}=8\)

=>|2x-6|=8

=>\(\left[{}\begin{matrix}2x-6=8\\2x-6=-8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=14\\2x=-2\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=7\left(nhận\right)\\x=-1\left(nhận\right)\end{matrix}\right.\)

e: ĐKXĐ: \(x\in R\)

\(\sqrt{4x^2+12x+9}=5\)

=>\(\sqrt{\left(2x\right)^2+2\cdot2x\cdot3+3^2}=5\)

=>\(\sqrt{\left(2x+3\right)^2}=5\)

=>|2x+3|=5

=>\(\left[{}\begin{matrix}2x+3=5\\2x+3=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=2\\2x=-8\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=1\left(nhận\right)\\x=-4\left(nhận\right)\end{matrix}\right.\)

f: ĐKXĐ:x>=6/5

\(\sqrt{5x-6}-3=0\)

=>\(\sqrt{5x-6}=3\)

=>\(5x-6=3^2=9\)

=>5x=6+9=15

=>x=15/5=3(nhận)

ĐKXĐ: x>=2 và y<>-x

\(\left\{{}\begin{matrix}\dfrac{4}{x+y}-2\sqrt{9x-18}=14\\\dfrac{5-2x-2y}{x+y}-\sqrt{4x-8}=-\dfrac{7}{2}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\dfrac{4}{x+y}-6\sqrt{x-2}=14\\\dfrac{5}{x+y}-2-2\sqrt{x-2}=-\dfrac{7}{2}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\dfrac{2}{x+y}-3\sqrt{x-2}=7\\\dfrac{5}{x+y}-2\sqrt{x-2}=-\dfrac{3}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{10}{x+y}-15\sqrt{x-2}=35\\\dfrac{10}{x+y}-4\sqrt{x-2}=-3\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}-11\sqrt{x-2}=38\left(vn\right)\\\dfrac{10}{x+y}-15\sqrt{x-2}=35\end{matrix}\right.\)

Vậy: Hệ vô nghiệm

NV
21 tháng 1

ĐKXĐ: \(x\ge2;x+y\ne0\)

\(\left\{{}\begin{matrix}\dfrac{4}{x+y}-6\sqrt{x-2}=14\\\dfrac{5-2\left(x+y\right)}{x+y}-2\sqrt{x-2}=-\dfrac{7}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2}{x+y}-3\sqrt{x-2}=7\\\dfrac{5}{x+y}-2-2\sqrt{x-2}=-\dfrac{7}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2}{x+y}-3\sqrt{x-2}=7\\\dfrac{5}{x+y}-2\sqrt{x-2}=-\dfrac{3}{2}\end{matrix}\right.\)

Đặt \(\left\{{}\begin{matrix}\dfrac{1}{x+y}=u\\\sqrt{x-2}=v\ge0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}2u-3v=7\\5u-2v=-\dfrac{3}{2}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}u=-\dfrac{37}{22}\\v=-\dfrac{38}{11}< 0\left(ktm\right)\end{matrix}\right.\)

Vậy hệ vô nghiệm

f) Ta có: \(\sqrt{16\left(x+1\right)}-\sqrt{9\left(x+1\right)}=4\)

\(\Leftrightarrow4\left|x+1\right|-3\left|x+1\right|=4\)

\(\Leftrightarrow\left|x+1\right|=4\)

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

g) Ta có: \(\sqrt{9x+9}+\sqrt{4x+4}=\sqrt{x+1}\)

\(\Leftrightarrow5\sqrt{x+1}-\sqrt{x+1}=0\)

\(\Leftrightarrow x+1=0\)

hay x=-1

NV
14 tháng 1 2021

1.

\(\Leftrightarrow\left(2x+1\right)\sqrt{2x^2+4x+5}-\left(2x+1\right)\left(x+3\right)+x^2-2x-4=0\)

\(\Leftrightarrow\left(2x+1\right)\left(\sqrt{2x^2+4x+5}-\left(x+3\right)\right)+x^2-2x-4=0\)

\(\Leftrightarrow\dfrac{\left(2x+1\right)\left(x^2-2x-4\right)}{\sqrt{2x^2+4x+5}+x+3}+x^2-2x-4=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2-2x-4=0\\\dfrac{2x+1}{\sqrt{2x^2+4x+5}+x+3}+1=0\left(1\right)\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow2x+1+\sqrt{2x^2+4x+5}+x+3=0\)

\(\Leftrightarrow\sqrt{2x^2+4x+5}=-3x-4\) \(\left(x\le-\dfrac{4}{3}\right)\)

\(\Leftrightarrow2x^2+4x+5=9x^2+24x+16\)

\(\Leftrightarrow7x^2+20x+11=0\)

NV
14 tháng 1 2021

2.

ĐKXĐ: ...

\(\Leftrightarrow2x\sqrt{2x+7}+7\sqrt{2x+7}=x^2+2x+7+7x\)

\(\Leftrightarrow\left(x^2-2x\sqrt{2x+7}+2x+7\right)+7\left(x-\sqrt{2x+7}\right)=0\)

\(\Leftrightarrow\left(x-\sqrt{2x+7}\right)^2+7\left(x-\sqrt{2x+7}\right)=0\)

\(\Leftrightarrow\left(x-\sqrt{2x+7}\right)\left(x+7-\sqrt{2x+7}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{2x+7}\\x+7=\sqrt{2x+7}\end{matrix}\right.\)

\(\Leftrightarrow...\)