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1 tháng 8 2018

1/

Ta có:  \(\left(1+\sqrt{15}\right)^2\)= 1 + 15 + \(2\sqrt{15}\)= 16 + \(2\sqrt{15}\)

              \(\sqrt{24}^2\)= 24 = 16 + 8

Vì:     \(\sqrt{15}^2\)= 15 < 16 =\(4^2\)

Nên:   \(\sqrt{15}< 4\)

=>       \(2\sqrt{15}< 8\)

=>       \(16+2\sqrt{15}< 24\)

=>      \(\left(1+\sqrt{15}\right)^2< \sqrt{24}^2\)

Vậy     \(1+\sqrt{15}< \sqrt{24}\)

2/

b/    \(3x-7\sqrt{x}=20\)\(\left(x\ge0\right)\)

<=> \(3x-7\sqrt{x}-20=0\)

<=> \(3x-12\sqrt{x}+5\sqrt{x}-20=0\)

<=> \(3\sqrt{x}\left(\sqrt{x}-4\right)+5\left(\sqrt{x}-4\right)=0\)

<=> \(\left(\sqrt{x}-4\right)\left(3\sqrt{x}+5\right)=0\)

<=> \(\sqrt{x}-4=0\)hoặc \(3\sqrt{x}+5=0\)

<=>   \(\sqrt{x}=4\)hoặc \(3\sqrt{x}=-5\)(vô nghiệm)

<=>   \(x=16\)

Vậy S=\(\left\{16\right\}\)

c/    \(1+\sqrt{3x}>3\)

<=> \(\sqrt{3x}>2\)

<=>   \(3x>4\)

<=>  \(x>\frac{4}{3}\)

d/      \(x^2-x\sqrt{x}-5x-\sqrt{x}-6=0\)(\(x\ge0\))

<=>   \(\left(x^2-5x-6\right)-\left(x\sqrt{x}+\sqrt{x}\right)=0\)

<=>   \(\left(x^2-6x+x-6\right)-\left(x\sqrt{x}+\sqrt{x}\right)=0\)

<=>    \([x\left(x-6\right)+\left(x-6\right)]-\sqrt{x}\left(x+1\right)=0\)

<=>   \(\left(x-6\right)\left(x+1\right)-\sqrt{x}\left(x+1\right)=0\)

<=>   \(\left(x+1\right)\left(x-6-\sqrt{x}\right)=0\)

<=>    \(\left(x+1\right)\left(x-3\sqrt{x}+2\sqrt{x}-6\right)=0\) 

<=>    \(\left(x+1\right)[\sqrt{x}\left(\sqrt{x}-3\right)+2\left(\sqrt{x}-3\right)]=0\)

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

<=>     \(x+1=0\)  hoặc \(\sqrt{x}-3=0\)hoặc \(\sqrt{x}+2=0\)

<=>     \(x=-1\)(loại)  hoặc \(x=9\)hoặc \(\sqrt{x}=-2\)(vô nghiệm)

Vậy S={  9 }

28 tháng 6 2017

đăng ít 1 thôi

10 tháng 9 2020

sao nhiều thế

a, \(16x^2-5=0\)

\(\Rightarrow16x^2=5\)

\(\Rightarrow x^2=\frac{5}{16}\)

\(\Rightarrow x=\sqrt{\frac{5}{16}}\Rightarrow x=\frac{\sqrt{5}}{4}\)

b, \(2\sqrt{x-3}=4\)

\(\Rightarrow\sqrt{x-3}=4:2\)

\(\Rightarrow\sqrt{x-3}=2\)

\(\Rightarrow x-3=4\)

\(\Rightarrow x=4+3\)

\(\Rightarrow x=7\)

c, \(\sqrt{4x^2-4x+1}=3\)

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

\(\Rightarrow2x-1=3\)

\(\Rightarrow2x=4\)

\(\Rightarrow x=2\)

d, \(\sqrt{x+3}\ge5\)

\(\Rightarrow x+3\ge25\)

\(\Rightarrow x\ge22\)

e, \(\sqrt{3x-1}< 2\)

\(\Rightarrow3x-1< 4\)

\(\Rightarrow3x< 5\)

\(\Rightarrow x< \frac{5}{3}\)

g, \(\sqrt{x^2-9}+\sqrt{x^2-6x+9}=0\)

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

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

\(\left(\sqrt{x+3}+\sqrt{x-3}\right)>0\)

\(\Rightarrow\sqrt{x-3}=0\)

\(\Rightarrow x-3=0\)

\(\Rightarrow x=3\)

7 tháng 7 2019

a) \(16x^2-5=0\)

\(\Leftrightarrow16x^2=5\)

\(\Leftrightarrow x^2=\frac{5}{16}\)

\(\Leftrightarrow x=\pm\sqrt{\frac{5}{16}}\)

b) \(2\sqrt{x-3}=4\)

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

\(\Leftrightarrow x-3=4\)

\(\Leftrightarrow x=7\)

c) \(\sqrt{4x^2-4x+1}=3\)

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

\(\Leftrightarrow2x-1=3\)

\(\Leftrightarrow2x=4\)

\(\Leftrightarrow x=2\)

d) \(\sqrt{x+3}\ge5\)

\(\Leftrightarrow x+3\ge25\)

\(\Leftrightarrow x\ge22\)

e) \(\sqrt{3x-1}< 2\)

\(\Leftrightarrow3x-1< 4\)

\(\Leftrightarrow3x< 5\)

\(\Leftrightarrow x< \frac{5}{3}\)

g) \(\sqrt{x^2-9}+\sqrt{x^2-6x+9}=0\)

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

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

Vì \(\left(\sqrt{x+3}+\sqrt{x-3}\right)>0\)

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

\(\Leftrightarrow x-3=0\)

\(\Leftrightarrow x=3\)

16 tháng 11 2019

a) b) c) bạn bình phương 2 vế

d) pt <=>3-x=x+3+2.căn(x+2)

<=> -2x=2.căn (x+2)

<=>-x=căn (x+2) (x<=0)

<=> x^2=x+2

<=>x=-1 hoặc x=2

Xong bạn xét ĐKXĐ

16 tháng 11 2019

giải giúp tớ a , b,c luôn đi cậu :<

a: \(\sqrt{x^2-4x+4}=3x+1\)

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

=>|x-2|=3x+1

=>\(\begin{cases}3x+1\ge0\\ \left(3x+1\right)^2=\left(x-2\right)^2\end{cases}\Rightarrow\begin{cases}x\ge-\frac13\\ \left(3x+1-x+2\right)\left(3x+1+x-2\right)=0\end{cases}\)

=>\(\begin{cases}x\ge-\frac13\\ \left(2x+3\right)\left(4x-1\right)=0\end{cases}\Rightarrow\begin{cases}x\ge-\frac13\\ x\in\left\lbrace-\frac32;\frac14\right\rbrace\end{cases}\)

=>\(x=\frac14\)

b:

ĐKXĐ: \(x^2-4x+1\ge0\)

=>\(x^2-4x+4-3\ge0\)

=>\(\left(x-2\right)^2\ge3\)

=>\(\left[\begin{array}{l}x-2\ge\sqrt3\\ x-2\le-\sqrt3\end{array}\right.\Rightarrow\left[\begin{array}{l}x\ge2+\sqrt3\\ x\le2-\sqrt3\end{array}\right.\)

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

=>\(\begin{cases}x\ge0\\ x^2-4x+1=x^2\end{cases}\Rightarrow\begin{cases}x\ge0\\ -4x+1=0\end{cases}\Rightarrow x=\frac14\)

c: \(\sqrt{x^2-2x+5}=x+3\)

=>\(\begin{cases}x+3\ge0\\ x^2-2x+5=\left(x+3\right)^2\end{cases}\Rightarrow\begin{cases}x\ge-3\\ x^2+6x+9=x^2-2x+5\end{cases}\)

=>\(\begin{cases}x\ge-3\\ x^2+6x+9-x^2+2x-5=0\end{cases}\Rightarrow\begin{cases}x\ge-3\\ 8x+4=0\end{cases}\Rightarrow x=-\frac12\)

d: \(\sqrt{x^2-10x+25}-2x=3\)

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

=>|x-5|=2x+3

=>\(\begin{cases}2x+3\ge0\\ \left(2x+3\right)^2=\left(x-5\right)^2\end{cases}\Rightarrow\begin{cases}x\ge-\frac32\\ \left(2x+3-x+5\right)\left(2x+3+x-5\right)=0\end{cases}\)

=>\(\begin{cases}x\ge-\frac32\\ \left(x+8\right)\left(3x-2\right)=0\end{cases}\Rightarrow x=\frac23\)

e:

ĐKXĐ: \(\left[\begin{array}{l}x\ge3\\ x\le1\end{array}\right.\)

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

=>\(\begin{cases}x-2\ge0\\ x^2-4x+3=\left(x-2\right)^2\end{cases}\Rightarrow\begin{cases}x\ge2\\ x^2-4x+3=x^2-4x+4\end{cases}\)

=>x∈∅

f: \(\sqrt{x^2-6x+9}=2x-1\)

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

=>|x-3|=2x-1

=>\(\begin{cases}2x-1\ge0\\ \left(2x-1\right)^2=\left(x-3\right)^2\end{cases}\Rightarrow\begin{cases}x\ge\frac12\\ \left(2x-1-x+3\right)\left(2x-1+x-3\right)=0\end{cases}\)

=>\(\begin{cases}x\ge\frac12\\ \left(x+2\right)\left(3x-4\right)=0\end{cases}\Rightarrow x=\frac43\)

26 tháng 6 2015

Rất muốn tính nhưng mà dài quá