Tìm min
a)|x+3|+|3x+5|+|4x+1|+5x+2
b)|2x+3|+|3x+4|+|4x+5|-6x+5
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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 $x=\frac{-5}{3}$
Tìm x biết
1. 2(5x-8)-3(4x-5)=4(3x-4)+11
2. (2x+1)2-(4x-1).(x-3)-15=0
3. (3x-1).(2x-7)-(1-3x).(6x-5)=0
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
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) 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=\(\dfrac{19}{24}\)