a: ĐKXĐ: \(x\in R\)

\(\sqrt{x^2-4x+4}=7\)

=>\(\sqrt{\left(x-2\right)^2}=7\)

=>|x-2|=7

=>\(\left[{}\begin{matrix}x-2=7\\x-2=-7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=9\\x=-5\end{matrix}\right.\)

b: ĐKXĐ: x>=-3

\(\sqrt{4x+12}-3\sqrt{x+3}+\dfrac{4}{3}\cdot\sqrt{9x+27}=6\)

=>\(2\sqrt{x+3}-3\sqrt{x+3}+\dfrac{4}{3}\cdot3\sqrt{x+3}=6\)

=>\(3\sqrt{x+3}=6\)

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

=>x+3=4

=>x=1(nhận)

6²=36;vay ta co 4x+12=36;4x=36-12;4x=24;vay x=6

THEO TÍNH CHẤT ,TA CÓ 

\(\sqrt{4x+12}=6\)

\(\Rightarrow4x+12=6^2\)

\(4x+12=36\)

\(4x=36-12\)

\(4x=24\)

\(x=24:4\)

\(x=6\)

x=6

k cho mình nha

\(\sqrt{4\left(1-x\right)^2}-6=0\) 

<=> \(\left|2\left(1-x\right)\right|=6\)

TH1: x \(\ge\)1 Khi đó pt trở thành:

\(2\left(x-1\right)=6\)

<=> x - 1 = 3

<=> x = 4 (tm)

TH2: x < 1, khi đó pt trở thành:

2(1 - x) = 6

<=> 1 - x = 3

<=> x = -2(tm)

vậy S= {4; -2}

Trả lời:

\(\sqrt{4\left(1-x\right)^2}-6=0\)

\(\Leftrightarrow2.\left|1-x\right|=6\)

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

\(\Leftrightarrow\orbr{\begin{cases}1-x=3\\1-x=-3\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-2\\x=4\end{cases}}\)

Vậy \(x=\left\{-2,4\right\}\)

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

\(\Leftrightarrow4x^2+4x+1=\left(x+2\right)^2\)

\(\Leftrightarrow4x^2+4x+1=x^2+4x+4\)

\(\Leftrightarrow3x^2=3\)

\(\Leftrightarrow x^2=1\)

\(\Leftrightarrow\orbr{\begin{cases}x=1\left(TM\right)\\x=-1\left(TM\right)\end{cases}}\)

Vậy \(x=\left\{1,-1\right\}\)

\(\sqrt{\sqrt{5}-\sqrt{\sqrt{3}-\sqrt{29-12\sqrt{5}}}}\)

\(=\sqrt{\sqrt{5}-\sqrt{\sqrt{3}-\sqrt{20-12\sqrt{5}+9}}}\)

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

\(=\sqrt{\sqrt{5}-\sqrt{\sqrt{3}-2\sqrt{5}+3}}\)

\(\sqrt{4x^2+4x+1}=x+2\) (Đk: x > = -2)

<=> \(\sqrt{\left(2x+1\right)^2}=x+2\)

<=>\(\left|2x+1\right|=x+2\)

<=> \(\orbr{\begin{cases}2x+1=x+2\left(đk:x\ge-\frac{1}{2}\right)\\-2x-1=x+2\left(đk:x\le-\frac{1}{2}\right)\end{cases}}\)

<=> \(\orbr{\begin{cases}x=1\left(tm\right)\\x=-1\left(tm\right)\end{cases}}\)

Vậy S = {1; -1}

a) \(\sqrt{4-5x}=12\)

ĐK : x ≤ 4/5

Bình phương hai vế

⇔ \(4-5x=144\)

⇔ \(-5x=140\)

⇔ \(x=-28\)( tm )

b) \(\sqrt{1-4x+4x^2}=5\)

⇔ \(\sqrt{\left(1-2x\right)^2}=5\)

⇔ \(\left|1-2x\right|=5\)

⇔ \(\orbr{\begin{cases}1-2x=5\\1-2x=-5\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-2\\x=3\end{cases}}\)

c) \(\sqrt{4x+20}-3\sqrt{5+x}+\frac{3}{4}\sqrt{9x+45}=6\)

ĐK : x ≥ -5

⇔ \(\sqrt{2^2\left(x+5\right)}-3\sqrt{x+5}+\frac{3}{4}\sqrt{3^2\left(x+5\right)}=6\)

⇔ \(\left|2\right|\sqrt{x+5}-3\sqrt{x+5}+\frac{3}{4}\cdot\left|3\right|\sqrt{x+5}=6\)

⇔ \(2\sqrt{x+5}-3\sqrt{x+5}+\frac{9}{4}\sqrt{x+5}=6\)

⇔ \(\frac{5}{4}\sqrt{x+5}=6\)

⇔ \(\sqrt{x+5}=\frac{24}{5}\)

⇔ \(x+5=\frac{576}{25}\)

⇔ \(x=\frac{451}{25}\)( tm )

d)\(\sqrt{x-2}\le3\)

ĐK : x ≥ 2

⇔ \(x-2\le9\)

⇔ \(x\le11\)

Kết hợp với điều kiện => Nghiệm của bpt là 2 ≤ x ≤ 11

\(a,\Leftrightarrow x-1=4\Leftrightarrow x=5\\ b,\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{3}{4}\\3x+1=4x-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{3}{4}\\x=4\left(tm\right)\end{matrix}\right.\Leftrightarrow x=4\\ c,ĐK:x\ge-5\\ PT\Leftrightarrow2\sqrt{x+5}-3\sqrt{x+5}+4\sqrt{x+5}=6\\ \Leftrightarrow3\sqrt{x+5}=6\\ \Leftrightarrow\sqrt{x+5}=3\\ \Leftrightarrow x+5=9\\ \Leftrightarrow x=4\left(tm\right)\)

\(d,\Leftrightarrow\sqrt{\left(x-2\right)^2}=\sqrt{\left(\sqrt{5}+1\right)^2}\\ \Leftrightarrow\left|x-2\right|=\sqrt{5}+1\\ \Leftrightarrow\left[{}\begin{matrix}x-2=\sqrt{5}+1\\2-x=\sqrt{5}+1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{5}+3\\x=1-\sqrt{5}\end{matrix}\right.\)

ĐKXĐ: x 0

√x - √(4x) + √(9x) = 6

√x - 2√x + 3√x = 6

2√x = 6

√x = 6 : 2

√x = 3

x = 9 (nhận)

Vậy x = 9

a) Ta có: \(\sqrt{4-5x}=12\)

\(\Leftrightarrow4-5x=144\)

\(\Leftrightarrow5x=-140\)

hay x=-28

b) Ta có: \(\sqrt{10+\sqrt{3x}}=2+\sqrt{6}\)

\(\Leftrightarrow\sqrt{3x}+10=10+4\sqrt{6}\)

\(\Leftrightarrow\sqrt{3x}=4\sqrt{6}\)

\(\Leftrightarrow3x=96\)

hay x=32

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

\(\Leftrightarrow2\sqrt{x+5}-3\sqrt{x+5}+\dfrac{4}{3}\cdot3\sqrt{x+5}=6\)

\(\Leftrightarrow x+5=4\)

hay x=-1

Mn giúp e với ak

a) \(\sqrt{x^2-6x+9}\)

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

\(=\sqrt{\left(x-3\right)^2}\) ≥0,∀x

⇒x∈\(R\)

b) \(\sqrt{x^2-2x+1}\)

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

\(=\sqrt{\left(x-1\right)^2}\) ≥0,∀x

⇒x∈\(R\)

a) \(\sqrt{x}>4\) có nghĩa là \(\sqrt{x}>\sqrt{16}\)

Vì \(x\ge0\) (x không âm) nên \(\sqrt{x}>\sqrt{16}\Leftrightarrow x>16\)

Vậy \(x>16\)

b) \(\sqrt{4x}\le4\) có nghĩa là \(\sqrt{4x}\le\sqrt{16}\)

Vì \(x\ge0\) (x không âm) nên \(\sqrt{4x}\le\sqrt{16}\Leftrightarrow4x\le16\Leftrightarrow x\le4\)

Vậy \(x\le4\)

c) \(\sqrt{4-x}\ge6\) có nghĩa là \(\sqrt{4-x}\ge\sqrt{36}\)

Vì \(x\ge0\) (x không âm) nên \(\sqrt{4-x}\ge\sqrt{36}\Leftrightarrow4-x\ge36\Leftrightarrow x\le-32\)

Vậy \(x\le-32\)

a) \(\sqrt{1-4x+4x^2}=5\)

<=> \(\sqrt{4x^2-4x+1}=5\)

<=> 4x² - 4x + 1 = 5²

<=> 4x² - 4x + 1 = 25

<=> 4x² - 4x + 1 - 25 = 0

<=> 4x² - 4x - 24 = 0

<=> 4(x + 2)(x - 3) = 0

<=> x = -2 hoặc x = 3

 => x = -2 hoặc x = 3

b) \(\sqrt{4-5x}=12\)

<=> \(\sqrt{-5x+4}=12\)

<=> -5x + 4 = 12²

<=> -5x + 4 = 144

<=> -5x = 144 - 4

<=> -5x = 140

<=> x = -28

=> x = -28

\(a,\sqrt{1-4x+4x^2}=5\)

\(\Rightarrow4x^2-4x+1=25\)

\(\Rightarrow4x^2-4x-24=0\)

\(\Rightarrow x^2-x-6=0\)

\(\Rightarrow x^2-3x+2x-6=0\)

\(\Rightarrow x\left(x-3\right)+2\left(x-3\right)=0\)

\(\Rightarrow\left(x-3\right)\left(x+2\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-3=0\\x+2=0\end{cases}\Rightarrow\orbr{\begin{cases}x=3\\x=-2\end{cases}}}\)

\(b,\sqrt{4-5x}=12\)

\(\Rightarrow4-5x=144\)

\(\Rightarrow5x=-140\)

\(\Rightarrow x=-28\)

\(a,\sqrt{1-4x+4x^2}=5\)

\(\Rightarrow\sqrt{1-4x+4x^2}=\sqrt{25}\)

\(\Rightarrow1-4x+4x^2=25\)

\(\Rightarrow4x^2-4x-24=0\)

\(\Rightarrow x^2-x-6=0\)

\(\Rightarrow x^2-3x+2x-6=0\)

\(\Rightarrow x\left(x-3\right)+2\left(x-3\right)=0\)

\(\Rightarrow\left(x-3\right)\left(x+2\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-3=0\\x+2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=3\\x=-2\end{cases}}\)

\(b,\sqrt{4-5x}=12\)

\(\Rightarrow\sqrt{4-5x}=\sqrt{144}\)

\(\Rightarrow4-5x=144\)

\(\Rightarrow x=-28\)

\(c,\sqrt{4x+20}-3\sqrt{5+x}+\frac{4}{3}\sqrt{9x+45}=6\)

\(\Rightarrow\sqrt{4\left(x+5\right)}-3\sqrt{x+5}+\frac{4}{3}\sqrt{9\left(x+5\right)}=6\)

\(\Rightarrow2\sqrt{x+5}-3\sqrt{x+5}+4\sqrt{x+5}=6\)

\(\Rightarrow\left(2-3+4\right)\sqrt{x+5}=6\)

\(\Rightarrow3\sqrt{x+5}=6\)

\(\Rightarrow\sqrt{x+5}=2\)

\(\Rightarrow\sqrt{x+5}=\sqrt{4}\)

\(\Rightarrow x+5=4\)

\(\Rightarrow x=-1\)