Cs vẻ bn đx qw rảnh rỗi òi -,-

4x² + 4x + 1 = 0

( 2x + 1 )² = 0

=> 2x + 1 = 0

=> x = -1/2

Hok tốt

đặt \(\sqrt{3x-2}=a\) và \(\sqrt{x-1}=b\)=> \(\sqrt{3x^2-5x+2}=ab\)

và \(4x=a^2+b^2+3\)

khi đó pt trên trở thành \(a+b=a^2+b^2+3+9+2ab\)

    đặt a+b=t thì pt trên trở thành \(t=12+t^2\)

                     <=> \(t^2-t+12=0\)

đến đây vô nghiệm rùi  nên cả pt vô nghiệm 

nk bạn mk nghĩ cái căn đầu tiên phải là \(\sqrt{3x-2}\) chứ

cái căn đầu là \(\sqrt{3x+2}\)

7: Ta có: \(\left(3x+4\right)\left(2x-1\right)+6x\left(1-x\right)=0\)

\(\Leftrightarrow6x^2-3x+8x-4+6x-6x^2=0\)

\(\Leftrightarrow11x=4\)

hay \(x=\dfrac{4}{11}\)

8: Ta có: \(2x\left(x^2-1\right)+x\left(-2x^2-3x+1\right)=-x-27\)

\(\Leftrightarrow2x^3-2x-2x^3-3x^2+x+x+27=0\)

\(\Leftrightarrow x^2=9\)

hay \(x\in\left\{3;-3\right\}\)

Ta có \(\frac{1}{\sqrt{1}}>\frac{1}{\sqrt{2}}>...\)\(>\frac{1}{\sqrt{n}}\)

Suy ra \(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+...+\)\(\frac{1}{\sqrt{n}}>\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{n}}\)\(+...+\frac{1}{\sqrt{n}}=n.\frac{1}{\sqrt{n}}=\sqrt{n}\)

a)√x−2+12√4x−8=√9x−18−2

=>√x−2+12√4(x−2)=√9(x−2)−2

=>√x−2+12√2²(x−2)=√3²(x−2)−2

=>√x−2+12.2√(x−2)=3√(x−2)−2

=>√x−2+24√(x−2)=3√(x−2)−2

=>√x−2+24√(x−2)-3√(x−2)=-2

=>√x−2(1+24-3)=-2

=>22√x−2=-2

=>√x−2=-2/22

=>√x−2=-1/11

=>x−2=1/121

=>x=1/121+2=243/121

b)√(3x−1)²=5

=>|3x−1|=5

=>3x−1=5 hoặc 3x−1=-5

=>3x=6 hoặc 3x=-4

=>x=2 hoặc x=-4/3

ĐKXĐ: \(\left\{{}\begin{matrix}2x+5>=0\\4-2x>=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}2x>=-5\\2x< =4\end{matrix}\right.\Leftrightarrow-\dfrac{5}{2}< =x< =2\)

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

=>\(x^2-4+\sqrt{2x+5}-3+\sqrt{4-2x}=4x-1-7\)

=>\(\left(x-2\right)\left(x+2\right)+\dfrac{2x+5-9}{\sqrt{2x+5}+3}+\sqrt{4-2x}=4x-8\)

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

=>\(-\left(2-x\right)\left[\left(x-2\right)+\dfrac{2}{\sqrt{2x+5}+3}\right]+\sqrt{2\left(2-x\right)}=0\)

=>\(\sqrt{2-x}\left[-\sqrt{2-x}\left(x-2+\dfrac{2}{\sqrt{2x+5}+3}\right)+\sqrt{2}\right]=0\)

=>\(\sqrt{2-x}=0\)

=>x=2(nhận)

Bài này sử dụng tính chất cơ bản: \(\left|A\right|\pm A\ge0\) với mọi A

a.

\(A=\left|-x-3\right|+\left|4x+1\right|+\left|3x+5\right|+5x+2\)

\(A\ge\left|3x-2\right|+\left|3x+5\right|+5x+2=\left|3x-2\right|+\dfrac{3}{2}.\left|2x+\dfrac{10}{3}\right|+5x+2\)

\(A\ge\left|3x-2\right|+\left|2x+\dfrac{10}{3}\right|+\dfrac{1}{2}\left|2x+\dfrac{10}{3}\right|+5x+2\)

\(A\ge\left|5x+\dfrac{4}{3}\right|+5x+\dfrac{4}{3}+\dfrac{1}{2}\left|2x+\dfrac{10}{3}\right|+\dfrac{2}{3}\ge\dfrac{2}{3}\)

\(A_{min}=\dfrac{2}{3}\) khi \(2x+\dfrac{10}{3}=0\Rightarrow x=-\dfrac{5}{3}\)

b. Tương tự

\(B\ge\left|5x+7\right|+\left|x+\dfrac{5}{4}\right|+3\left|x+\dfrac{5}{4}\right|-6x+5\)

\(B\ge\left|6x+\dfrac{33}{4}\right|-\left(6x+\dfrac{33}{4}\right)+3\left|x+\dfrac{5}{4}\right|+\dfrac{53}{4}\ge\dfrac{53}{4}\)

\(B_{min}=\dfrac{53}{4}\) khi \(x=-\dfrac{5}{4}\)

Lời giải:

a. Áp dụng BĐT $|a|+|b|\geq |a+b|$ ta có:

\(A=|-x-3|+|4x+1|+|3x+5|+5x+2\)

\(\geq |-x-3+4x+1|+|3x+5|+5x+2=|3x-2|+|3x+5|+5x+2\)

Nếu $x\geq \frac{2}{3}$ thì:

$A\geq 3x-2+3x+5+5x+2=11x+5\geq 11.\frac{2}{3}+5=\frac{37}{3}$

Nếu $\frac{-5}{3}\leq x< \frac{2}{3}$ thì:

$A\geq 2-3x+3x+5+5x+2=9+5x\geq 9+5.\frac{-5}{3}=\frac{2}{3}$

Nếu $x< \frac{-5}{3}$ thì:

$A\geq 2-3x-3x-5+5x+2=-1-x>\frac{2}{3}$

Từ 3 TH trên suy ra $A_{\min}=\frac{2}{3}$ khi $x=\frac{-5}{3}$

b. Áp dụng BĐT $|a|+|b|\geq |a+b|$ thì:

\(B\geq |2x+3+3x+4|+|4x+5|-6x+5=|5x+7|+|4x+5|-6x+5\)

Nếu $x\geq \frac{-5}{4}$ thì:

$B\geq 5x+7+4x+5-6x+5=3x+17\geq 3.\frac{-5}{4}+17=\frac{53}{4}$ 

Nếu $\frac{-7}{5}\leq x< \frac{-5}{4}$ thì:

$B\geq 5x+7-4x-5-6x+5=-5x+7> -5.\frac{-5}{4}+7=\frac{53}{4}$

Nếu $x< \frac{-7}{5}$ thì:

$B\geq -5x-7-4x-5-6x+5=-15x-7> -15.\frac{-7}{5}-7=14$

Từ 3 TH trên suy ra $B_{\min}=\frac{53}{4}$. Giá trị này đạt tại $x=\frac{-5}{4}$