Rút gon :
a, A = 2x - 3,5 - | 4x + 4,8 | - 3x + 1
b, B = - | - 4 - 2,5x | + 4x - 6,2 - 5,2x
c, C = 2,2x - 6 + | 4,2 - 1,4 x | - 3x +1,2
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a, \(A=-x-2,5-\left|4x+4,8\right|\)
\(b,B=-\left|-4-2,5x\right|-\dfrac{6}{5}x-6,2\)
\(c,C=\left|4,2-1,4x\right|-\dfrac{4}{5}x-\dfrac{24}{5}\)
19 22 25 28 5(3x + 2) – 4(2x +3) x*(1 + 2x) 4(1 + x) – 3(2x-5) 4x–8(6) - X) 23/ ... 2x” – 4x + 3x – 6 = 2x” – X-6 (b) (x-3) = (x-3)(x-3) (c) (2x+y)(2x–y) = x* = x* – 3x ... (x - 6)” 7 (3x + 5)(x-6) 8 (8x + 2)(3x + 4) (4x – 1)(2x – 3) 10 (2x +5)* 11 (8x – 3)(2x + ... 27 (4x + 3y)(x + y) 28 (2x + 5)(5x – 2) (4x – 3y)(4x + y) 30 (7x + 2y)(3x + 4y) 24/ ...
\(a)=3x\cdot\left(2x-7-4x+5\right)=3x\cdot\left(-2x-2\right)=3x\cdot\left[-2\cdot\left(x+1\right)\right]\)
#)Giải :
b) Với : x < -6 , phương trình có dạng :
- x - 6 = 2x + 9
<=> -3x = 15
<=> x = - 5 ( không thỏa mãn )
Với : x ≥ - 6 , phương trình có dạng :
x + 6 = 2x + 9
<=> x = - 3 ( thỏa mãn)
Vậy , phương trình nhận : x = - 3 làm nghiệm duy nhất
#~Will~be~Pens~#
A, \(\left|9+6x\right|=2x\Rightarrow\orbr{\begin{cases}9+6x=2x\\9+6x=-2x\end{cases}\Rightarrow\orbr{\begin{cases}2x-6x=9\\-2x-6x=9\end{cases}}}\Rightarrow\orbr{\begin{cases}-4x=9\\-8x=9\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{-4}{9}\\x=\frac{-8}{9}\end{cases}}\)
B, \(\left|x+6\right|=2x+9\Rightarrow\orbr{\begin{cases}x+6=2x+9\\x+6=-2x-9\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x-2x=9-6\\x+2x=-9-6\end{cases}}\Rightarrow\orbr{\begin{cases}-x=3\\3x=-15\end{cases}}\Rightarrow\orbr{\begin{cases}x=-3\\x=-5\end{cases}}\)
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
a: \(A=2x-3.5-3x+1-\left|4x+4.8\right|\)
\(=-x-2.5-\left|4x+4.8\right|\)
Trường hợp 1: x>=-1,2
=>A=-x-2,5-4x-4,8=-5x-7,3
Trường hợp 2: x<-1,2
=>A=-x-2,5+4x+4,8=3x+2,3
b: \(B=-\left|2.5x+4\right|-1.2x-6.2\)
Trường hợp 1: x>=-1,6
=>B=-2,5x-4-1,2x-6,2=-3,7x-10,2
Trường hợp 2: x<-1,6
B=2,5x+4-1,2x-6,2=1,3x-2,2
c: \(C=-0.8x-4.8+\left|1.4x-4.2\right|\)
Trường hợp 1: x>=3
C=-0,8x-4,8+1,4x-4,2=0,6x-9
Trường hợp 2: x<3
C=-0,8x-4.8-1.4x+4,2=-2,2x-0,6