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1 tháng 9 2023

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 tháng 9 2023

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) \(\sqrt{\left(x-3\right)^2}=3-x\)

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

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

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

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

\(\Leftrightarrow x=3\)

14 tháng 10 2018

\(\sqrt{x+6}+\sqrt{x-3}-\sqrt{x+1}-\sqrt{x-2}=0\)(ĐKXĐ: \(x\ge3\))

\(\Leftrightarrow\sqrt{x+6}+\sqrt{x-3}=\sqrt{x+1}+\sqrt{x-2}\)

\(\Leftrightarrow2x+3+2\sqrt{\left(x-3\right)\left(x+6\right)}=2x-1+2\sqrt{\left(x+1\right)\left(x-2\right)}\)

\(\Leftrightarrow2+\sqrt{\left(x-3\right)\left(x+6\right)}=\sqrt{\left(x+1\right)\left(x-2\right)}\)

\(\Leftrightarrow\left(2+\sqrt{\left(x-3\right)\left(x+6\right)}\right)^2=x^2-x-2\)

\(\Leftrightarrow x^2+3x-14+4\sqrt{\left(x-3\right)\left(x+6\right)}=x^2-x-2\)

\(\Leftrightarrow4x-12+4\sqrt{\left(x-3\right)\left(x+6\right)}=0\)

\(\Leftrightarrow\left(x-3\right)+\sqrt{\left(x-3\right)\left(x+6\right)}=0\)

\(\Leftrightarrow\sqrt{x-3}\left(\sqrt{x-3}+\sqrt{x+6}\right)=0\)\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x-3}=0\\\sqrt{x-3}+\sqrt{x+6}=0\end{cases}}\)

+) Nếu \(\sqrt{x-3}=0\Leftrightarrow x-3=0\Leftrightarrow x=3\)

+) Nếu \(\sqrt{x-3}+\sqrt{x+6}=0\). Ta thấy: \(\hept{\begin{cases}\sqrt{x-3}\ge0\\\sqrt{x+6}\ge0\end{cases}}\forall x\in R\)

Do đó: \(\hept{\begin{cases}\sqrt{x-3}=0\\\sqrt{x+6}=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=3\\x=-6\end{cases}}\)\(\Rightarrow x=3\)(loại \(x=-6\) vì không t/m ĐKXĐ)

Vậy pt có một nghiệm duy nhất là x= 3.

16 tháng 10 2021

a. \(\sqrt[3]{1-2x}+3=0\left(ĐK:x\le\dfrac{1}{2}\right)\)

<=> \(\sqrt[3]{1-2x}=-3\)

<=> \(1-2x=\left(-3\right)^3\)

<=> \(1-2x=-27\)

<=> \(-2x=-28\)

<=> \(x=14\left(TM\right)\)

b: Ta có: \(\sqrt{x^2-6x+9}-\dfrac{\sqrt{6}+\sqrt{3}}{\sqrt{2}+1}=0\)

\(\Leftrightarrow x^2-6x+9=3\)

\(\Leftrightarrow x^2-6x+6=0\)

\(\text{Δ}=\left(-6\right)^2-4\cdot1\cdot6=36-24=12\)

Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:

\(\left\{{}\begin{matrix}x_1=\dfrac{6-2\sqrt{3}}{2}=3-\sqrt{3}\\x_2=3+\sqrt{3}\end{matrix}\right.\)

18 tháng 5 2021

b)đk:\(x\ge\dfrac{1}{2}\)

Có: \(\sqrt{2x^2-1}\le\dfrac{2x^2-1+1}{2}=x^2\)

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

=>\(\sqrt{2x^2-1}+x\sqrt{2x-1}\le2x^2\) 

Dấu = xảy ra\(\Leftrightarrow x=1\)

Vậy....

c) đk: \(x\ge0\)

\(\Leftrightarrow\sqrt{x}=\sqrt{x+9}-\dfrac{2\sqrt{2}}{\sqrt{x+1}}\)
\(\Rightarrow x=x+9+\dfrac{8}{x+1}-4\sqrt{\dfrac{2\left(x+9\right)}{x+1}}\)

\(\Leftrightarrow0=9+\dfrac{8}{x+1}-4\sqrt{\dfrac{2\left(x+9\right)}{x+1}}\)

Đặt \(a=\sqrt{\dfrac{2\left(x+9\right)}{x+1}}\left(a>0\right)\)

\(\Leftrightarrow\dfrac{a^2-2}{2}=\dfrac{8}{x+1}\)

pttt \(9+\dfrac{a^2-2}{2}-4a=0\) \(\Leftrightarrow a=4\) (TM)

\(\Rightarrow4=\sqrt{\dfrac{2\left(x+9\right)}{x+1}}\) \(\Leftrightarrow16=\dfrac{2\left(x+9\right)}{x+1}\) \(\Leftrightarrow x=\dfrac{1}{7}\) (TM)
Vậy ...

 

18 tháng 5 2021

a)ĐKXĐ: x≥-1/3; x≤6

<=>\(\dfrac{3x-15}{\sqrt{3x+1}+4}+\dfrac{x-5}{\sqrt{x-6}+1}+\left(x-5\right)\cdot\left(3x+1\right)=0\Leftrightarrow\left(x-5\right)\cdot\left(\dfrac{3}{\sqrt{3x+1}+4}+\dfrac{1}{\sqrt{x-6}+1}+3x+1\right)=0\Leftrightarrow x-5=0\Leftrightarrow x=5\)(nhận)

(vì x≥-1/3 nên3x+1≥0 )

 

c: \(\Leftrightarrow x-3=0\)

hay x=3

4 tháng 1 2022

c: ⇔x−3=0⇔x−3=0

hay x=3

NV
6 tháng 8 2021

1.

ĐKXĐ: \(x< 5\)

\(\Leftrightarrow\sqrt{\dfrac{42}{5-x}}-3+\sqrt{\dfrac{60}{7-x}}-3=0\)

\(\Leftrightarrow\dfrac{\dfrac{42}{5-x}-9}{\sqrt{\dfrac{42}{5-x}}+3}+\dfrac{\dfrac{60}{7-x}-9}{\sqrt{\dfrac{60}{7-x}}+3}=0\)

\(\Leftrightarrow\dfrac{9x-3}{\left(5-x\right)\left(\sqrt{\dfrac{42}{5-x}}+3\right)}+\dfrac{9x-3}{\left(7-x\right)\left(\sqrt{\dfrac{60}{7-x}}+3\right)}=0\)

\(\Leftrightarrow\left(9x-3\right)\left(\dfrac{1}{\left(5-x\right)\left(\sqrt{\dfrac{42}{5-x}}+3\right)}+\dfrac{1}{\left(7-x\right)\left(\sqrt{\dfrac{60}{7-x}}+3\right)}\right)=0\)

\(\Leftrightarrow x=\dfrac{1}{3}\)

NV
6 tháng 8 2021

b.

ĐKXĐ: \(x\ge2\)

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

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

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

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

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

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

\(\Rightarrow x=2\)