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a, \(\left|x+2\right|-\left|x+7\right|=0\Rightarrow\left|x+2\right|=\left|x+7\right|\Rightarrow\orbr{\begin{cases}x+2=x+7\\x+2=-x-7\end{cases}\Rightarrow\orbr{\begin{cases}0=5\left(loại\right)\\2x=-9\end{cases}\Rightarrow}x=\frac{-9}{2}}\)
b, - Nếu \(2x-1\ge0\Rightarrow x\ge\frac{1}{2}\), ta có: 2x - 1 = 2x - 1 => 2x = 2x (thỏa mãn với mọi x)
- Nếu 2x - 1 < 0 => \(x< \frac{1}{2}\), ta có: 2x - 1 = 1 - 2x => 4x = 2 => x = \(\frac{1}{2}\) (không thỏa mãn điều kiện)
Vậy \(x\ge\frac{1}{2}\)
c,d tương tự b
e, tương tự a
a.
\(\left(3x-1\right)\left(2x+7\right)-\left(x+1\right)\left(6x-5\right)=16\)
\(6x^2+21x-2x-7-6x^2+5x-6x+5=16\)
\(\left(6x^2-6x^2\right)+\left(21x-2x+5x-6x\right)-\left(7-5\right)=16\)
\(18x-2=16\)
\(18x=16+2\)
\(18x=18\)
\(x=\frac{18}{18}\)
\(x=1\)
b.
\(\left(10x+9\right)x-\left(5x-1\right)\left(2x+3\right)=8\)
\(10x^2+9x-10x^2-15x+2x+3=8\)
\(\left(10x^2-10x^2\right)-\left(15x-9x-2x\right)+3=8\)
\(-4x=8-3\)
\(-4x=5\)
\(x=-\frac{5}{4}\)
c.
\(\left(3x-5\right)\left(7-5x\right)+\left(5x+2\right)\left(3x-2\right)-2=0\)
\(21x-15x^2-35+25x+15x^2-10x+6x-4-2=0\)
\(\left(15x^2-15x^2\right)+\left(25x+21x-10x+6x\right)-\left(35+4+2\right)=0\)
\(42x=41\)
\(x=\frac{41}{42}\)
=>(2x+3).(10x+2)=(5x+2).(4x+5)
=>(2x.10x)+(2x.2)+(3.10x)+(3.2)=(5x.4x)+(5x.5)+(2.4x)+(2.5)
=>20x2+4x+30x+6=20x2+25x+8x+10
=>20x2-20x2+4x-8x+30x-25x=10-6
=>0+4x-8x+30x-25x=4
=>-4x+30x-25x=4
=>26x-25x=4
=>x=4
B)=>(3x-1).(5x-34)=(40-5x).(25-3x)
=>15x2-102x-5x+34=1000-120x-125x+15x2
=>15x2-107x+34=1000-245x+15x2
=>15x2-15x2-107x+245x=1000-34
=>0-107x+245x=966
=>138x=966
=>x=7
A,=>(2x+3).(10x+2)=(5x+2).(4x+5)
=>(2x.10x)+(2x.2)+(3.10x)+(3.2)=(5x.4x)+(5x.5)+(2.4x)+(2.5)
=>20x2+4x+30x+6=20x2+25x+8x+10
=>20x2-20x2+4x-8x+30x-25x=10-6
=>0+4x-8x+30x-25x=4
=>-4x+30x-25x=4
=>26x-25x=4
=>x=4
Bài 1:\(\left|2x-1\right|=2x-1\) khi \(x>0\)
b)\(\left|0,5-3x\right|=3x-0.5\) khi x= 4
c)\(\left|5x+1\right|-10x=0,5\) khi x= 0,1
Bài 2:Min A=0
Min B=-2
Bài 1:
a, \(\left|2x-1\right|=2x-1\)
+) Xét \(x\ge\dfrac{1}{2}\) ta có:
\(2x-1=2x-1\)
\(\Rightarrow x\) tùy ý với \(x\ge\dfrac{1}{2}\)
+) Xét \(x< \dfrac{1}{2}\) ta có:
\(1-2x=2x-1\)
\(\Rightarrow4x=2\)
\(\Rightarrow x=\dfrac{1}{2}\) ( không t/m )
Vậy...
b, \(\left|0,5-3x\right|=3x-0,5\)
+) Xét \(x\ge\dfrac{1}{6}\) ta có:
\(0,5-3x=3x-0,5\)
\(\Rightarrow6x=1\)
\(\Rightarrow x=\dfrac{1}{6}\) ( t/m )
+) Xét \(x< \dfrac{1}{6}\) ta có:
\(3x-0,5=3x-0,5\)
\(\Rightarrow x\) tùy ý với \(x< \dfrac{1}{6}\)
Vậy \(x\le\dfrac{1}{6}\)
c, \(\left|5x+1\right|-10x=0,5\)
+) Xét \(x\ge\dfrac{-1}{5}\) ta có:
\(5x+1-10x=0,5\)
\(\Rightarrow-5x=-0,5\)
\(\Rightarrow x=\dfrac{1}{10}\) ( t/m )
+) Xét \(x< \dfrac{-1}{5}\) ta có:
\(-5x-1-10x=0,5\)
\(\Rightarrow-15x=1,5\)
\(\Rightarrow x=\dfrac{-1}{10}\) ( không t/m )
Vậy \(x=\dfrac{1}{10}\)
Bài 2:
a, Ta có: \(-\left|x-3,5\right|\le0\)
\(\Rightarrow A=0,5-\left|x-3,5\right|\le3,5\)
Dấu " = " xảy ra khi \(-\left|x-3,5\right|=0\Rightarrow x=3,5\)
Vậy \(MIN_A=0,5\) khi x = 3,5
b, Ta có: \(-\left|1,4-x\right|\le0\)
\(\Rightarrow B=-\left|1,4-x\right|-2\le-2\)
Dấu " = " xảy ra khi \(-\left|1,4-x\right|=0\Rightarrow x=1,4\)
Vậy \(MIN_B=-2\) khi \(x=1,4\)
a. 5x.(12x + 7) - 3x(20x - 5) = - 100
↔ 60x2 + 35x - 60x2 + 15x = - 100
↔ 50x = - 100
→ x = - 2
b. 0,6x(x - 0,5) - 0,3x(2x + 1,3) = 0,138
↔ 0,6x2 - 0,3x - 0,6x2 - 0,39x = 0,138
↔ - 0,6x = 0,138
↔ x = 0,138 : (- 0,6)
↔ x = - 0,2
Bài 4. Tìm x biết:
a) 5x(12x + 7) – 3x(20x – 5) = - 100
\(\Leftrightarrow60x^2+35x-60x^2+15x=-100\)
\(\Leftrightarrow50x=-100\)
\(\Leftrightarrow x=-2\)
b) 0,6x(x – 0,5) – 0,3x(2x + 1,3) = 0,138
\(\Leftrightarrow0,6x^2-0,3x-0,6x^2-0,39x=0,138\)
\(\Leftrightarrow-0,69x=0,138\)
\(\Leftrightarrow x=-0,2\)
c) 6x(5x + 3) + 3x(1 – 10x) = 7
\(\Leftrightarrow30x^2+18x+3x-30x^2=7\)
\(\Leftrightarrow21x=7\)
\(\Leftrightarrow x=\frac{1}{3}\)
Bài 2:
3x + 2(5 - x) = 0
<=> 3x + 10 - 2x = 0
<=> x + 10 = 0
<=> x = 0 - 10
<=> x = -10
=> x = -10
Bài 3:
6(3q + 4q) - 8(5p - q) + (p - q)
= 6.3p + 6.4q - 8.5p - (-8).q + p - q
= 18p + 24q - 40p + 8q + p - q
= (18p - 40p + p) + (24q + 8q - q)
= -21p + 31q
`@` `\text {Ans}`
`\downarrow`
`a)`
`3x(4x-1) - 2x(6x-3) = 30`
`=> 12x^2 - 3x - 12x^2 + 6x = 30`
`=> 3x = 30`
`=> x = 30 \div 3`
`=> x=10`
Vậy, `x=10`
`b)`
`2x(3-2x) + 2x(2x-1) = 15`
`=> 6x- 4x^2 + 4x^2 - 2x = 15`
`=> 4x = 15`
`=> x = 15/4`
Vậy, `x=15/4`
`c)`
`(5x-2)(4x-1) + (10x+3)(2x-1) = 1`
`=> 5x(4x-1) - 2(4x-1) + 10x(2x-1) + 3(2x-1)=1`
`=> 20x^2-5x - 8x + 2 + 20x^2 - 10x +6x - 3 =1`
`=> 40x^2 -17x - 1 = 1`
`d)`
`(x+2)(x+2)-(x-3)(x+1)=9`
`=> x^2 + 2x + 2x + 4 - x^2 - x + 3x + 3=9`
`=> 6x + 7 =9`
`=> 6x = 2`
`=> x=2/6 =1/3`
Vậy, `x=1/3`
`e)`
`(4x+1)(6x-3) = 7 + (3x-2)(8x+9)`
`=> 24x^2 - 12x + 6x - 3 = 7 + (3x-2)(8x+9)`
`=> 24x^2 - 12x + 6x - 3 = 7 + 24x^2 +11x - 18`
`=> 24x^2 - 6x - 3 = 24x^2 + 18x -11`
`=> 24x^2 - 6x - 3 - 24x^2 + 18x + 11 = 0`
`=> 12x +8 = 0`
`=> 12x = -8`
`=> x= -8/12 = -2/3`
Vậy, `x=-2/3`
`g)`
`(10x+2)(4x- 1)- (8x -3)(5x+2) =14`
`=> 40x^2 - 10x + 8x - 2 - 40x^2 - 16x + 15x + 6 = 14`
`=> -3x + 4 =14`
`=> -3x = 10`
`=> x= - 10/3`
Vậy, `x=-10/3`