mong mọi người giải giúp em vs 

1) \(\sqrt[]{9\left(x-1\right)}=21\)

\(\Leftrightarrow9\left(x-1\right)=21^2\)

\(\Leftrightarrow9\left(x-1\right)=441\)

\(\Leftrightarrow x-1=49\Leftrightarrow x=50\)

2) \(\sqrt[]{1-x}+\sqrt[]{4-4x}-\dfrac{1}{3}\sqrt[]{16-16x}+5=0\)

\(\Leftrightarrow\sqrt[]{1-x}+\sqrt[]{4\left(1-x\right)}-\dfrac{1}{3}\sqrt[]{16\left(1-x\right)}+5=0\)

\(\)\(\Leftrightarrow\sqrt[]{1-x}+2\sqrt[]{1-x}-\dfrac{4}{3}\sqrt[]{1-x}+5=0\)

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

\(\Leftrightarrow\sqrt[]{1-x}.\dfrac{8}{3}=-5\)

\(\Leftrightarrow\sqrt[]{1-x}=-\dfrac{15}{8}\)

mà \(\sqrt[]{1-x}\ge0\)

\(\Leftrightarrow pt.vô.nghiệm\)

3) \(\sqrt[]{2x}-\sqrt[]{50}=0\)

\(\Leftrightarrow\sqrt[]{2x}=\sqrt[]{50}\)

\(\Leftrightarrow2x=50\Leftrightarrow x=25\)

1) \(\sqrt{9\left(x-1\right)}=21\) (ĐK: \(x\ge1\))

\(\Leftrightarrow3\sqrt{x-1}=21\)

\(\Leftrightarrow\sqrt{x-1}=7\)

\(\Leftrightarrow x-1=49\)

\(\Leftrightarrow x=49+1\)

\(\Leftrightarrow x=50\left(tm\right)\)

2) \(\sqrt{1-x}+\sqrt{4-4x}-\dfrac{1}{3}\sqrt{16-16x}+5=0\) (ĐK: \(x\le1\))

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

\(\Leftrightarrow\dfrac{5}{3}\sqrt{1-x}+5=0\)

\(\Leftrightarrow\dfrac{5}{3}\sqrt{1-x}=-5\) (vô lý) 

Phương trình vô nghiệm

3) \(\sqrt{2x}-\sqrt{50}=0\) (ĐK: \(x\ge0\)) 

\(\Leftrightarrow\sqrt{2x}=\sqrt{50}\)

\(\Leftrightarrow2x=50\)

\(\Leftrightarrow x=\dfrac{50}{2}\)

\(\Leftrightarrow x=25\left(tm\right)\)

4) \(\sqrt{4x^2+4x+1}=6\)

\(\Leftrightarrow\sqrt{\left(2x+1\right)^2}=6\)

\(\Leftrightarrow\left|2x+1\right|=6\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+1=6\left(ĐK:x\ge-\dfrac{1}{2}\right)\\2x+1=-6\left(ĐK:x< -\dfrac{1}{2}\right)\end{matrix}\right.\)

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

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

5) \(\sqrt{\left(x-3\right)^2}=3-x\)

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

\(\Leftrightarrow x-3=3-x\)

\(\Leftrightarrow x+x=3+3\)

\(\Leftrightarrow x=\dfrac{6}{2}\)

\(\Leftrightarrow x=3\)

1) => 9(x-1)=\(21^2\)

=> 9x-9=441

=> 9x=450

=> x=50

2)=>\(\sqrt{1-x}\) + \(\sqrt{4\left(1-x\right)}\)-\(\dfrac{1}{3}\sqrt{16\left(1-x\right)}\)+5=0

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

=>\(\dfrac{5}{3}\sqrt{1-x}\) +5=0

=>\(\sqrt{1-x}\)=-3

Phuong trinh vo nghiem

b. Tự đặt đk

\(x^{^2}+5\sqrt{x-3}=21\\\Leftrightarrow x^{^2}-9+5\sqrt{x-3}=12 \)

Đặt \(a=\sqrt{x-3}\) \(\left(a\ge0\right)\) Phương trình trở thành:

\(a^{^2}\left(a^{^2}+6\right)+5a=12\\ \Leftrightarrow a^{^4}+6a^{^2}+5a-12=0\\ \Leftrightarrow a^{^4}-a^{^3}+a^{^3}-a^{^2}+7a^{^2}-7a+12a-12=0\\ \Leftrightarrow\left(a-1\right)\left(a^{^3}+a^{^2}+7a+12\right)=0\\ \Leftrightarrow a=1\left(tmdk\right)\)

Ta có: vì \(a\ge0\) nên \(a^{^3}+a^{^2}+7a+12\ne0\)

Với a = 1 ta có x=4 (tmdk)

a, ĐKXĐ : \(x\ge\dfrac{1}{2}\)

 PT <=> 2x - 1 = 5

<=> x = 3 ( TM )

Vậy ...

b, ĐKXĐ : \(x\ge5\)

PT <=> x - 5 = 9

<=> x = 14 ( TM )

Vậy ...

c, PT <=> \(\left|2x+1\right|=6\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+1=6\\2x+1=-6\end{matrix}\right.\)

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

Vậy ...

d, PT<=> \(\left|x-3\right|=3-x\)

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

Vậy phương trình có vô số nghiệm với mọi x \(x\le3\)

e, ĐKXĐ : \(-\dfrac{5}{2}\le x\le1\)

PT <=> 2x + 5 = 1 - x

<=> 3x = -4

<=> \(x=-\dfrac{4}{3}\left(TM\right)\)

Vậy ...

f ĐKXĐ : \(\left[{}\begin{matrix}x\le0\\1\le x\le3\end{matrix}\right.\)

PT <=> \(x^2-x=3-x\)

\(\Leftrightarrow x=\pm\sqrt{3}\) ( TM )

Vậy ...

a) \(\sqrt{2x-1}=\sqrt{5}\)          (x \(\ge\dfrac{1}{2}\))

<=> 2x - 1 = 5

<=> x = 3 (tmđk)

Vậy S = \(\left\{3\right\}\)

b) \(\sqrt{x-5}=3\)           (x\(\ge5\))

<=> x - 5 = 9

<=> x = 4 (ko tmđk)

Vậy x \(\in\varnothing\)

c) \(\sqrt{4x^2+4x+1}=6\)          (x \(\in R\))

<=> \(\sqrt{\left(2x+1\right)^2}=6\)

<=> |2x + 1| = 6

<=> \(\left[{}\begin{matrix}\text{2x + 1=6}\\\text{2x + 1}=-6\end{matrix}\right.< =>\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=\dfrac{-7}{2}\end{matrix}\right.\)(tmđk)

Vậy S = \(\left\{\dfrac{5}{2};\dfrac{-7}{2}\right\}\)

a.\(2x^2+5x+8+\sqrt{x}=x^2+3x+35+x^2+2x-7\)

\(=2x^2+5x+8+\sqrt{x}=2x^2+5x+28\Leftrightarrow\sqrt{x}=20\Leftrightarrow x=400.\)

b.\(3\sqrt{x}+7x+5=\sqrt{x}+4x-6+3x+18\)

\(=3\sqrt{x}+7x+5=\sqrt{x}+7x+12\Leftrightarrow2\sqrt{x}=7\Leftrightarrow x=\frac{49}{4}.\)

c.\(8\sqrt{x}+2x-9=5x+7+6\sqrt{x}-3x-12.\)

\(=8\sqrt{x}+2x-9=2x+6\sqrt{x}-5\Leftrightarrow2\sqrt{x}=4\Leftrightarrow x=4.\)

d.\(2\sqrt{3x}+11x-18=5x+3+6\sqrt{3x}+6x-21\)

\(=2\sqrt{3x}+11x-18=11x+6\sqrt{3x}-19\Leftrightarrow4\sqrt{3x}=1\)

\(\Leftrightarrow\sqrt{3x}=\frac{1}{4}\Leftrightarrow3x=\frac{1}{16}\Leftrightarrow x=\frac{1}{48}.\)

a) \(2x^2+5x+8+\sqrt{x}=x^2+3x+35+x^2+2x-7\)

<=> \(2x^2+5x+8+\sqrt{x}=2x^2+5x+28\)

<=> \(2x^2+5x+8+\sqrt{x}-\left(2x^2+5\right)=28\)

<=> \(\sqrt{x}+8=28\)

<=> \(\sqrt{x}=28-8\)

<=> \(\sqrt{x}=20\)

<=> \(\left(\sqrt{x}\right)^2=20^2\)

<=> x = 400

=> x = 400

b) \(3\sqrt{x}+7x+5=\sqrt{x}+4x-6+3x+18\)

<=> \(3\sqrt{x}+7x+5=7x+\sqrt{x}+12\)

<=> \(3\sqrt{x}+5=7x+\sqrt{x}+12-7x\)

<=> \(3\sqrt{x}+5=\sqrt{x}+12\)

<=> \(3\sqrt{x}=\sqrt{x}+12-5\)

<=> \(3\sqrt{x}=\sqrt{x}+7\)

<=> \(3\sqrt{x}-\sqrt{x}=7\)

<=> \(2\sqrt{x}=7\)

<=> \(\sqrt{x}=\frac{7}{2}\)

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

<=> \(x=\frac{49}{4}\)

=> \(x=\frac{49}{4}\)

c) \(8\sqrt{x}+2x-9=5x+7+6\sqrt{x}-3x-12\)

<=> \(8\sqrt{x}+2x-9=2x+6\sqrt{x}-5\)

<=> \(8\sqrt{x}-9=2x+6\sqrt{x}-5-2x\)

<=> \(8\sqrt{x}-9=6\sqrt{x}-5\)

<=> \(8\sqrt{x}=6\sqrt{x}-5+9\)

<=> \(8\sqrt{x}=6\sqrt{x}+4\)

<=> \(8\sqrt{x}-6\sqrt{x}=4\)

<=> \(2\sqrt{x}=4\)

<=> \(\sqrt{x}=2\)

<=> \(\left(\sqrt{x}\right)^2=2^2\)

<=> x = 4

=> x = 4

d) \(2\sqrt{3x}+11x-18=5x+3+6\sqrt{3x}+6x-21\)

<=> \(2\sqrt{3x}+11x-18=11x+6\sqrt{3x}-18\)

<=> \(2\sqrt{3x}+11x-18-\left(11x-18\right)=6\sqrt{3x}\)

<=>\(2\sqrt{3x}=6\sqrt{3x}\)

<=> \(2\sqrt{3x}-6\sqrt{3x}=0\)

<=>\(-4\sqrt{3x}=0\)

<=> \(\sqrt{3x}=0\)

<=> \(\left(\sqrt{3x}\right)^2=0^2\)

<=> 3x = 0

<=> x = 0

=> x = 0

à, phần a ra x = 400. Nhầm

a/ ĐKXĐ:...

\(\Leftrightarrow4x^2-4x\sqrt{2x-1}-3x^2+6x-3=0\)

\(\Leftrightarrow4x\left(x-\sqrt{2x-1}\right)-3\left(x-1\right)^2=0\)

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

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

\(\left(1\right)\Leftrightarrow4x=3x+3\sqrt{2x-1}\)

\(\Leftrightarrow x=3\sqrt{2x-1}\)

\(\Leftrightarrow x^2-18x+9=0\) \(\Rightarrow9\pm6\sqrt{2}\)

Vậy pt có 3 nghiệm....

b/ ĐKXĐ:...

\(\Leftrightarrow4x^2-4x\sqrt{4x-3}-x^2+4x-3=0\)

\(\Leftrightarrow4x\left(x-\sqrt{4x-3}\right)-\left(x^2-4x+3\right)=0\)

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

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

\(\left(1\right)\Leftrightarrow4x=x+\sqrt{4x-3}\)

\(\Leftrightarrow3x=\sqrt{4x-3}\)

\(\Leftrightarrow9x^2-4x+3=0\) (vô nghiệm)

Vậy...

c/

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

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

Đặt \(\left\{{}\begin{matrix}x+1=a\\\sqrt{x^2+3}=b\end{matrix}\right.\)

\(\Rightarrow a^2+2b^2-3ab=0\)

\(\Leftrightarrow\left(a-b\right)\left(a-2b\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}a=b\\a=2b\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2+3}=x+1\left(x\ge-1\right)\\2\sqrt{x^2+3}=x+1\left(x\ge-1\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2+3=x^2+2x+1\\4x^2+12=x^2+2x+1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=2\\3x^2-2x+11=0\left(vn\right)\end{matrix}\right.\) \(\Rightarrow x=1\)