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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 $x=\frac{-5}{4}$

1) \(\Rightarrow10x-16-12x+15=12x-16+11\)

\(\Rightarrow14x=4\Rightarrow x=\dfrac{2}{7}\)

2) \(\Rightarrow4x^2+4x+1-4x^2+13x-3-15=0\)

\(\Rightarrow17x=17\Rightarrow x=1\)

3) \(\Rightarrow\left(3x-1\right)\left(2x-7+6x-5\right)=0\)

\(\Rightarrow\left(2x-3\right)\left(3x-1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{1}{3}\end{matrix}\right.\)

2: Ta có: \(\left(2x+1\right)^2-\left(4x-1\right)\left(x-3\right)-15=0\)

\(\Leftrightarrow4x^2+4x+1-4x^2+12x+x-3-15=0\)

\(\Leftrightarrow17x=17\)

hay x=1

* Trả lời:

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

\(\Leftrightarrow-3+6x-4-12x=-5x+5\)

\(\Leftrightarrow6x-12x+5x=3+4+5\)

\(\Leftrightarrow x=12\)

\(\left(2\right)\) \(3\left(2x-5\right)-6\left(1-4x\right)=-3x+7\)

\(\Leftrightarrow6x-15-6+24x=-3x+7\)

\(\Leftrightarrow6x+24x+3x=15+6+7\)

\(\Leftrightarrow33x=28\)

\(\Leftrightarrow x=\dfrac{28}{33}\)

\(\left(3\right)\) \(\left(1-3x\right)-2\left(3x-6\right)=-4x-5\)

\(\Leftrightarrow1-3x-6x+12=-4x-5\)

\(\Leftrightarrow-3x-6x+4x=-1-12-5\)

\(\Leftrightarrow-5x=-18\)

\(\Leftrightarrow x=\dfrac{18}{5}\)

\(\left(4\right)\) \(x\left(4x-3\right)-2x\left(2x-1\right)=5x-7\)

\(\Leftrightarrow4x^2-3x-4x^2+2x=5x-7\)

\(\Leftrightarrow-x-5x=-7\)

\(\Leftrightarrow-6x=-7\)

\(\Leftrightarrow x=\dfrac{7}{6}\)

\(\left(5\right)\) \(3x\left(2x-1\right)-6x\left(x+2\right)=-3x+4\)

\(\Leftrightarrow6x^2-3x-6x^2-12x=-3x+4\)

\(\Leftrightarrow-15x+3x=4\)

\(\Leftrightarrow-12x=4\)

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

1: Ta có: \(\left(x+3\right)^2-\left(x+2\right)\left(x-2\right)=4x+17\)

\(\Leftrightarrow x^2+6x+9-x^2+4-4x=17\)

\(\Leftrightarrow x=2\)

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

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

\(\Leftrightarrow6x=6\)

hay x=1

Noob ơi, bạn phải đưa vào máy tính ý solve cái là ra x luôn, chỉ tội là đợi hơi lâu

a, 4.(18 - 5x) - 12(3x - 7) = 15(2x - 16) - 6(x + 14) 

=> 72 - 20x - 36x + 84 = 30x - 240 - 6x - 84

=> (72 + 84) + (-20x - 36x) = (30x - 6x) + (-240 - 84) 

=> 156 -  56x = 24x - 324 

=>  24x + 56x = 324 + 156 

=> 80x = 480 

=> x = 480 : 80 =  6 

Vậy x = 6 

b, 5(3x + 5) - 4(2x - 3) = 5x + 3(2x + 12) + 1 

=> 15x + 25 - 8x + 12 = 5x + 6x + 36 + 1 

=> (15x - 8x) + (25 + 12) = 11x + 37 

=> 7x + 37 = 11x + 37 

=> 11x - 7x = 0 

=>  x = 0 

Rút gọn hết ta được :

a/ 41x - 17 = -21

=> 41x = -4 => x = 4/41

b/ 34x - 17 = 0 

=> 34x = 17

=> x = 17/34 = 1/2

c/ 19x + 56 = 52 

=> 19x = -4

=> x = -4/19

d/ 20x² - 16x - 34 = 10x² + 3x - 34

=> 10x² - 19x = 0

=> x(10x - 19) = 0

=> x = 0 

hoặc 10x - 19 = 0 => 10x = 19 => x = 19/10

Vậy x = 0 ; x = 19/10

Rút gọn hết ta được :

a/ 41x - 17 = -21

=> 41x = -4 => x = 4/41

b/ 34x - 17 = 0

=> 34x = 17

=> x = 17/34 = 1/2

c/ 19x + 56 = 52

=> 19x = -4

=> x = -4/19

d/ 20x 2 - 16x - 34 = 10x 2 + 3x - 34

=> 10x 2 - 19x = 0

=> x(10x - 19) = 0

=> x = 0 hoặc 10x - 19 = 0

=> 10x = 19

=> x = 19/10

Vậy x = 0 ; x = 19/10 

a) ( 6x - 3 ) ( 2x + 4 ) + ( 4x - 1 ) ( 5 - 3x ) = -21

<=> 12x² + 24x - 6x - 12 + 20x - 12x² - 5 + 3x = -21

<=> 41x = -21 + 12 + 5 

<=> 41x = -4

<=> x = -4/41

a) 3x(4x-3)-2x(5-6x)=0

\(\Leftrightarrow12x^2-9x-10x+12x^2=0\)

\(\Leftrightarrow24x^2-19x=0\)

\(\Leftrightarrow x\left(24x-19\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\24x-19=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\24x=19\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{19}{24}\end{matrix}\right.\)

Vậy x=0 hoặc x=\(\dfrac{19}{24}\)

b) 5(2x-3)+4x(x-2)+2x(3-2x)=0

\(\Leftrightarrow\)10x-15+4x²-8x+6x-4x²=0

\(\Leftrightarrow8x-15=0\)

\(\Leftrightarrow8x=15\)

\(\Leftrightarrow x=\dfrac{15}{8}\)

vậy x=\(\dfrac{15}{8}\)

c)3x(2-x)+2x(x-1)=5x(x+3)

\(\Leftrightarrow6x-3x^2+2x^2-2x=5x^2+15x\\ \Leftrightarrow4x-x^2=5x^2+15x\\ \Leftrightarrow4x-x^2-5x^2-15x=0\\ \)

\(\Leftrightarrow-6x^2-11x=0\\ \Leftrightarrow-x\left(6x+11\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}-x=0\\6x+11=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\6x=-11\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{-11}{6}\end{matrix}\right.\)

Vậy x=0 hoặc x=\(\dfrac{-11}{6}\)