Question

a. \(\sqrt{\left(2x+3\right)^2}=x+1\)

b. \(\sqrt{\left(2x-1\right)^2}=x+1\)

c. \(\sqrt{x+3}=5\)

d. \(\sqrt{x+2}=\sqrt{7}\) 

e. \(5\sqrt{x}=20\) 

f. \(\sqrt{x+4}=7\) 

g. \(\sqrt{\left(2x+1\right)^2}=3\)

Answer 1

a, \(\sqrt{\left(2x+3\right)^2}=x+1\)

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

TH1: \(\left\{{}\begin{matrix}2x+3=x+1\\2x+3\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-2\\x\ge-\dfrac{3}{2}\end{matrix}\right.\Rightarrow\) vô nghiệm.

Vậy phương trình vô nghiệm.

TH2: \(\left\{{}\begin{matrix}-2x-3=x+1\\2x+3< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{4}{3}\\x< -\dfrac{3}{2}\end{matrix}\right.\Rightarrow\) vô nghiệm.

Answer 2

b, 

a, \(\sqrt{\left(2x-1\right)^2}=x+1\)

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

TH1: \(\left\{{}\begin{matrix}2x-1=x+1\\2x-1\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\x\ge\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow x=2\)

TH2: \(\left\{{}\begin{matrix}-2x+1=x+1\\2x-1< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\x< \dfrac{1}{2}\end{matrix}\right.\Leftrightarrow x=0\)

Answer 3

tìm x, biết

 

Answer 4

c, ĐK: \(x\ge-3\)

\(\sqrt{x+3}=5\)

\(\Leftrightarrow x+3=25\)

\(\Leftrightarrow x=22\left(tm\right)\)

d, ĐK: \(x\ge-2\)

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

\(\Leftrightarrow x+2=7\)

\(\Leftrightarrow x=5\left(tm\right)\)

Answer 5

e, ĐK: \(x\ge0\)

\(5\sqrt{x}=20\)

\(\Leftrightarrow\sqrt{x}=4\)

\(\Leftrightarrow x=16\left(tm\right)\)

f, ĐK: \(x\ge-4\)

\(\sqrt{x+4}=7\)

\(\Leftrightarrow x+4=49\)

\(\Leftrightarrow x=45\left(tm\right)\)

Answer 6

g, \(\sqrt{\left(2x+1\right)^2}=3\)

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

\(\Leftrightarrow2x+1=\pm3\)

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

Answer 7

c. \(\sqrt{x+3}=5\)

<=> x + 3 = 5²

<=> x = 25 - 3

<=> x = 22

Answer 8

f. \(\sqrt{x+4}=7\)

<=> x + 4 = 7²

<=> x = 49 - 4

<=> x = 45

Answer 9

e. 5\(\sqrt{x}\) = 20

<=> \(\sqrt{x}\) = 20 : 5

<=> \(\sqrt{x}\) = 4

<=> x = 4² = 16

Answer 10

g. \(\sqrt{\left(2x+1\right)^2}\) = 3

<=> 2x + 1 = 3

<=> 2x = 2

<=> x = 1

Answer 11

a. \(\sqrt{\left(2x+3\right)^2}\) = x + 1

<=> 2x + 3 = x + 1

<=> 2x - x = 1 - 3

<=> x = -2

Answer 12

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

<=> 2x - 1 = x + 1

<=> 2x - x = 1 + 1

<=> x = 2

Answer 13

d. \(\sqrt{x+2}=\sqrt{7}\)

<=> \(x+2=\left(\sqrt{7}\right)^2\)

<=> x + 2 = 7

<=> x = 7 - 2

<=> x = 5

Answer 14

a: ta có: \(\sqrt{\left(2x+3\right)^2}=x+1\)

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

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}2x-1=x+1\left(x\ge\dfrac{1}{2}\right)\\2x-1=-x-1\left(x< \dfrac{1}{2}\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\left(nhận\right)\\x=0\left(nhận\right)\end{matrix}\right.\)

c: Ta có: \(\sqrt{x+3}=5\)

\(\Leftrightarrow x+3=25\)

hay x=22