tìm x
a) (2x-3)2=(4-5x)
b) 5.(x-1)3=20.(x-1)
c)9x2-16=3x-4
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a, 2x - 3 = 4 - 5x
=> 2x + 5x = 4 + 3
=> 7x = 7
=> x = 7 : 7
=> x = 1
b) 5.(x - 1)3 = 20.(x - 1)
=> 5.x3 - 3x3 + 3x - 1 = 20x - 1
=> x3(5 - 3) + 3x = 20x - 1 + 1
=> 2x3 + 3x = 20x
=> 2x3 = 20x - 3x
=> 2.x3 = 17x
=> 2.x3 - 17x = 0
=> x(2x2 - 17) = 0
=> x = 0 hoặc 2x2 - 17 = 0
=> 2x2 = 17
=> x2 = 8,5
=> x = \(\sqrt{8,5}\)
c) 9x2 - 16 = 3x - 4
=> 9x2 - 3x = -4 + 16
=> x(9x - 3) = 12
=> x[3(3x - 1)] = 12
=> x(3x - 1) = 12 : 3
=> x(3x - 1) = 4 chju nhe
to rat ngu
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
1) Ta có: \(5\left(x-3\right)\left(x-7\right)-\left(5x+1\right)\left(x-2\right)=-8\)
\(\Leftrightarrow5\left(x^2-10x+21\right)-\left(5x^2-10x+x-2\right)=-8\)
\(\Leftrightarrow5x^2-50x+105-5x^2+9x+2+8=0\)
\(\Leftrightarrow-41x=-115\)
hay \(x=\dfrac{115}{41}\)
2) Ta có: \(x\left(x+1\right)\left(x+2\right)-\left(x+4\right)\left(3x-5\right)=84-5x\)
\(\Leftrightarrow x\left(x^2+3x+2\right)-\left(3x^2+7x-20\right)=84-5x\)
\(\Leftrightarrow x^3+3x^2+2x-3x^2-7x+20-84+5x=0\)
\(\Leftrightarrow x^3=64\)
hay x=4
3) Ta có: \(\left(9x^2-5\right)\left(x+3\right)-3x^2\left(3x+9\right)=\left(x-5\right)\left(x+4\right)-x\left(x-11\right)\)
\(\Leftrightarrow9x^3+27x^2-5x-15-9x^3-27x^2=x^2-x-20-x^2+11x\)
\(\Leftrightarrow-5x-15=10x-20\)
\(\Leftrightarrow-5x-10x=-20+15\)
\(\Leftrightarrow x=\dfrac{-5}{-15}=\dfrac{1}{3}\)
a, \(4\left(18-5x\right)-12\left(3x-7\right)=15\left(2x-16\right)-6\left(x+14\right)\)
\(\Leftrightarrow72-20x-36x+84=30x-240-6x-84\)
\(\Leftrightarrow156-56x=24x-324\)
\(\Leftrightarrow-80x+480=0\Leftrightarrow x=-6\)
b, \(5\left(3x+5\right)-4\left(2x-3\right)=5x+3\left(2x-12\right)+1\)
\(\Leftrightarrow15x+25-8x+12=5x+6x-36+1\)
\(\Leftrightarrow7x+37=11x-35\)
\(\Leftrightarrow-4x+72=0\Leftrightarrow x=18\)
c, \(2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\)
\(\Leftrightarrow10x-16-12x+15=12x-16+11\)
\(\Leftrightarrow-2x-1=12x-5\)
\(\Leftrightarrow-14x+4=0\Leftrightarrow x=\frac{2}{7}\)
d, \(5x-3\left\{4x-2\left[4x-3\left(5x-2\right)\right]\right\}=182\)
\(\Leftrightarrow5x-3\left[4x-15x+6\right]=182\)
\(\Leftrightarrow5x-3\left(-11x+6\right)=182\)
\(\Leftrightarrow5x+33x-18-182=0\)
\(\Leftrightarrow38x-200=0\Leftrightarrow x=\frac{100}{19}\)
a: 3x-5>15-x
=>4x>20
hay x>5
b: \(3\left(x-2\right)\left(x+2\right)< 3x^2+x\)
=>3x2+x>3x2-12
=>x>-12
Bài 1:
a) (3x - 2)(4x + 5) = 0
<=> 3x - 2 = 0 hoặc 4x + 5 = 0
<=> 3x = 2 hoặc 4x = -5
<=> x = 2/3 hoặc x = -5/4
b) (2,3x - 6,9)(0,1x + 2) = 0
<=> 2,3x - 6,9 = 0 hoặc 0,1x + 2 = 0
<=> 2,3x = 6,9 hoặc 0,1x = -2
<=> x = 3 hoặc x = -20
c) (4x + 2)(x^2 + 1) = 0
<=> 4x + 2 = 0 hoặc x^2 + 1 # 0
<=> 4x = -2
<=> x = -2/4 = -1/2
d) (2x + 7)(x - 5)(5x + 1) = 0
<=> 2x + 7 = 0 hoặc x - 5 = 0 hoặc 5x + 1 = 0
<=> 2x = -7 hoặc x = 5 hoặc 5x = -1
<=> x = -7/2 hoặc x = 5 hoặc x = -1/5
a: \(\Leftrightarrow4x^2-12x+9-4+5x=0\)
\(\Leftrightarrow4x^2-7x+5=0\)
hay \(x\in\varnothing\)
b: =>(x-1)3-4(x-1)=0
=>(x-1)(x+1)(x-3)=0
hay \(x\in\left\{1;-1;3\right\}\)
c: =>(3x-4)(3x+4)-(3x-4)=0
=>(3x-4)(3x+3)=0
=>x=4/3 hoặc x=-1