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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
1) a) \(\left(3x-1\right)\left(9x^2+3x+1\right)-4x\left(x-5\right)\)
\(=27x^3+9x^2+3x-9x^2-3x-1-4x^2+20x\)
\(=27x^3+\left(9x^2-9x^2-4x^2\right)+\left(3x-3x+20x\right)+\left(-1\right)\)
\(=27x^3-4x^2+20x-1\)
b)\(\left(7x+2\right)\left(3-4x\right)-\left(x+3\right)\left(x^2-3x+9\right)\)
\(=21x-28x^2+6-8x-x^3+3x^2-9x-3x^2+9x-27\)
\(=\left(21x-8x-9x+9x\right)+\left(-28x^2+3x^2-3x^2\right)\)\(+\left(6-27\right)\)\(+\left(-x^3\right)\)
\(=13x-28x^2-21-x^3\)
c)\(\left(4x+3\right)\left(4x-3\right)-\left(2-x\right)\left(4+2x+x^2\right)\)
\(=16x^2-12x+12x-9-8-4x-2x^2+4x+2x^2+x^3\)
\(=\left(16x^2-2x^2+2x^2\right)+\left(-12x+12x-4x+4x\right)\)\(+\left(-9-8\right)\)\(+x^3\)
\(=16x^2-17+x^3\)
d)\(\left(3x-8\right)\left(-5x+6\right)-\left(4x+1\right)\left(3x-2\right)\)
\(=-15x^2+18x+40x-48-12x^2+8x-3x+2\)
\(=\left(-15x^2-12x^2\right)+\left(18x+40x+8x-3x\right)\)\(+\left(-48+2\right)\)
\(=-27x^2+63x-46\)
e)\(\left(3x-6\right)4x-2x\left(3x+5\right)-4x^2\)
\(=12x^2-24x-6x^2-10x-4x^2\)
\(=\left(12x^2-6x^2-4x^2\right)+\left(-24x-10x\right)\)
\(=2x^2-34x\)
f)\(\left(5x-6\right)\left(6x-5\right)-x\left(3x+10\right)\)
\(=30x^2-25x-36x+30-3x^2-10x\)
\(=\left(30x^2-3x^2\right)+\left(-25x-36x-10x\right)+30\)
\(=27x^2-71x+30\)
2) a)\(x\left(x+3\right)-x^2=6\)
\(\Rightarrow x^2+3x-x^2=6\)
\(\Rightarrow\left(x^2-x^2\right)+3x=6\)
\(\Rightarrow3x=6\)
\(\Rightarrow x=2\)
Vậy x=2
b) \(2x\left(x-5\right)+x\left(-2x-1\right)=6\)
\(\Rightarrow2x^2-10x-2x^2-x=6\)
\(\Rightarrow\left(2x^2-2x^2\right)+\left(-10x-x\right)=6\)
\(\Rightarrow-11x=6\)
\(\Rightarrow x=-\dfrac{6}{11}\)
\(\)Vậy \(x=-\dfrac{6}{11}\)
c) x(x+5)-(x+1)(x-2)=7
\(\Rightarrow x^2+5x-x^2+2x-x+2=7\)
\(\Rightarrow\left(x^2-x^2\right)+\left(5x+2x-x\right)=7-2\)
\(\Rightarrow6x=5\)
\(\Rightarrow x=\dfrac{5}{6}\)
Vậy x=\(\dfrac{5}{6}\)
d)\(\left(3x+4\right)\left(6x-3\right)-\left(2x+1\right)\left(9x-2\right)=10\)
\(\Rightarrow18x^2-9x+24x-12-18x^2+4x-9x+2=10\)
\(\Rightarrow\left(18x^2-18x^2\right)+\left(-9x+24x+4x-9x\right)+\left(-12+2\right)=10\)
\(\Rightarrow10x-10=10\)
\(\Rightarrow10x=20\)
\(\Rightarrow x=2\)
Vậy x=2
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
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
\(o,x^2-9x+20=0\)
\(\Leftrightarrow x^2-4x-5x+20=0\)
\(\Leftrightarrow x\left(x-4\right)-5\left(x-4\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(x-5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-4=0\\x-5=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=4\\x=5\end{cases}}\)
\(n,3x^3-3x^2-6x=0\)
\(\Leftrightarrow3x\left(x^2-x-2\right)=0\)
\(\Leftrightarrow3x\left(x^2+x-2x-2\right)=0\)
\(\Leftrightarrow3x\left[x\left(x+1\right)-2\left(x+1\right)\right]=0\)
\(\Leftrightarrow3x\left(x+1\right)\left(x-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}\orbr{\begin{cases}3x=0\\x+1=0\end{cases}}\\x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}\orbr{\begin{cases}x=0\\x=-1\end{cases}}\\x=2\end{cases}}\)
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 $x=\frac{-5}{3}$