Tìm x
(x+3)3-x(3x+1)2+(2x+1)(4x2-2x+1)=28
Giải chi tiết giúp mình vs nhé!
Thanks bạn
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\(a,\frac{1}{2}x+\frac{5}{2}=\frac{7}{2}x-\frac{3}{4}\)
\(\Leftrightarrow\frac{1}{2}x+\frac{5}{2}-\frac{7}{2}x=-\frac{3}{4}\)
\(\Leftrightarrow\frac{1}{2}x-\frac{7}{2}x+\frac{5}{2}=-\frac{3}{4}\)
\(\Leftrightarrow-3x+\frac{5}{2}=-\frac{3}{4}\)
\(\Leftrightarrow-3x=-\frac{13}{4}\)
\(\Leftrightarrow x=-\frac{13}{4}:(-3)=-\frac{13}{4}:\frac{-3}{1}=-\frac{13}{4}\cdot\frac{-1}{3}=\frac{13}{12}\)
\(b,\frac{2}{3}x-\frac{2}{5}=\frac{1}{2}x-\frac{1}{3}\)
\(\Leftrightarrow\frac{2}{3}x-\frac{2}{5}-\frac{1}{2}x=-\frac{1}{3}\)
\(\Leftrightarrow\frac{2}{3}x-\frac{1}{2}x-\frac{2}{5}=-\frac{1}{3}\)
\(\Leftrightarrow\frac{1}{6}x-\frac{2}{5}=-\frac{1}{3}\)
\(\Leftrightarrow\frac{1}{6}x=\frac{1}{15}\)
\(\Leftrightarrow x=\frac{1}{15}:\frac{1}{6}=\frac{1}{15}\cdot6=\frac{6}{15}=\frac{2}{5}\)
\(c,\frac{1}{3}x+\frac{2}{5}(x+1)=0\)
\(\Leftrightarrow\frac{1}{3}x+\frac{2}{5}x+\frac{2}{5}=0\)
\(\Leftrightarrow\frac{11}{15}x=-\frac{2}{5}\)
\(\Leftrightarrow x=-\frac{6}{11}\)
d,e,f Tương tự
Vd1:
d) Ta có: \(\sqrt{2}\left(x-1\right)-\sqrt{50}=0\)
\(\Leftrightarrow\sqrt{2}\left(x-1-5\right)=0\)
\(\Leftrightarrow x=6\)
\(\dfrac{2\text{x}-1}{3}=\dfrac{3\text{x}+1}{4}\)
\(\Leftrightarrow=\dfrac{4\left(2\text{x}-1\right)}{12}=\dfrac{3\left(3\text{x}+1\right)}{12}\)
\(\Leftrightarrow8\text{x}-4=9\text{x}+3\)
\(\Leftrightarrow8\text{x}-9\text{x}=3+4\)
\(\Leftrightarrow-x=7\)
\(\Leftrightarrow x=-7\)
a) 6x(3x +5)-2x(9x-2)=17
6x3x+6x5-2x9x-2x(-2)=17
\(18x^2\)+30x-\(18x^2\)+4x=17
\(18x^2-18x^2\)+ 34x=17
0 +34x=17
x=17:34
x=0.5
b)2x(3x-1)-3x(2x+11)-70=0
2x3x-2x1-3x2x+3x11-70=0
\(6x^2-2x-6x^2+33x-70=0\)
-2x+33x-70=0
31x-70=0
31x=0+70
31x=70
x=\(\frac{70}{31}\)
(trong câu c dấu . của mình là nhân nha)
c)5x(2x-3)-4(8-3x)=2(3+5x)
5x2x-5x3-4.8+4.3x=2.3+2.5x
\(10x^2-15x-32+12x=6+10x\)
\(10x^2-15x+12x-10x=6+32\)
\(10x^2-13x=38\)
tạm thời mình bí chổ này thông cảm nha bạn
a) (x + 2) . (x + 3) - (x - 2) . (x + 5) = 6
=> (x . x + 3x + 2x + 2 . 3) - (x . x + 5x - 2x - 2 . 5) = 6
=> (x2 + 5x + 6) - (x2 + 3x - 10) = 6
=> x2 + 5x + 6 - x2 - 3x + 10 = 6
=> 2x +16 = 6 => 2x = -10 => x = -5
b) (3x + 2) . (2x + 9) - (x + 2) . (6x + 1) = (x + 1) - (x - 6)
=> (3x . 2x + 3x . 9 + 2 . 2x + 2 . 9) - (x . 6x + 1x + 2 . 6x + 2 .1) = x + 1 - x + 6
=> (6x2 + 31x + 18) - (6x2 + 13x + 2) = 7
=> 6x2 + 31x + 18 - 6x2 - 13x - 2 = 7
=> 18x + 16 = 7 => 18x = 9 => x = 0,5
c) 3 . (2x - 1) . (3x - 1) - (2x - 3) . (9x - 1) = 0
=> 3(2x . 3x - 2x -3x + 1) - (2x . 9x - 2x -3 . 9x + 3) = 0
=> 3(6x2 - 5x +1) - (18x2 - 29x + 3) = 0
=> (18x2 -15x + 1) -(18x2 - 29x +3) = 0
=> 18x2 - 15x +1 -18x2 + 29x - 3 = 0
=> 14x = 0 => x = 0
a)(x+2)(x+3)-(x-2)(x+5)=6
x(x+3)+2(x+3)-x(x+5)+2(x+5)=6
x2+3x+2x+6-x2-5x+2x+10=6
(x2-x2)+(3x+2x-5x+2x)+(10+6)=6
2x+16=6
2x=6-16
2x=-10
x=-10/2
x=-5. Vậy x=-5
b)3x(2x+9)+2(2x+9)-x(6x+1)-2(6x+1)=x+1-x+6
6x2+27x+4x+18-6x2-x-12x-2=7
(6x2-6x2)+(27x+4x-x-12x)+(18-2)=7
18x+16=7
18x=7-16
x=-9/18=-1/2. Vậy x=-1/2
c)[3(3x-1)](2x-1)-(2x-3)(9x-1)=0
(9x-3)(2x-1)-(2x-3)(9x-1)=0
9x(2x-1)-3(2x-1)-2x(9x-1)+3(9x-1)=0
18x2-9x-6x+3-18x2+2x+27x-3=0
(18x2-18x2)+(27x+2x-6x-9x)+(3-3)=0
14x=0
x=0/14
x=0. Vậy x=0
a) (x + 2) . (x + 3) - (x - 2) . (x + 5) = 6 => (x . x + 3x + 2x + 2 . 3) - (x . x + 5x - 2x - 2 . 5) = 6
=> (x2 + 5x + 6) - (x2 + 3x - 10) = 6
=> x2 + 5x + 6 - x2 - 3x + 10 = 6
=> 2x +16 = 6 => 2x = -10 => x = -5
b) (3x + 2) . (2x + 9) - (x + 2) . (6x + 1) = (x + 1) - (x - 6)
=> (3x . 2x + 3x . 9 + 2 . 2x + 2 . 9) - (x . 6x + 1x + 2 . 6x + 2 .1) = x + 1 - x + 6
=> (6x2 + 31x + 18) - (6x2 + 13x + 2) = 7
=> 6x2 + 31x + 18 - 6x2 - 13x - 2 = 7
=> 18x + 16 = 7 => 18x = -9 => x = -0,5
c) 3 . (2x - 1) . (3x - 1) - (2x - 3) . (9x - 1) = 0
=> 3(2x . 3x - 2x - 3x + 1) - (2x . 9x - 2x - 3. 9x + 3) = 0
=> 3(6x2 - 5x + 1) - (18x2 - 29x + 3) = 0
=> 18x2 - 15x + 3 - 18x2 + 29x -3 = 0
=> 14x = 0 => x = 0.
Cho g(x) = 0
x + 1 = 0
x = -1
Để f(x) chia hết cho g(x) thì x = -1 cũng là nghiệm của f(x)
Hay f(1) = 0
3.1² + 2.1² - 7.1 - m + 2 = 0
-2 - m + 2 = 0
m = 0
Vậy m = 0 thì f(x) chia hết cho g(x)
Giải chi tiết của em đây :
F(x) = 3x2 + 2x2 - 7x - m + 2
F(x) \(⋮\) x + 1 \(\Leftrightarrow\) F(x) \(⋮\) x - (-1)
Theo bezout ta có : F(x) \(⋮\) x - (-1) \(\Leftrightarrow\) F(-1) = 0
\(\Leftrightarrow\) 3(-1)2 + 2(-1)2 - 7.(-1) - m + 2 = 0
3 + 2 + 7 - m + 2 =0
14 - m = 0
m = 14
Kết luận với m = 14 thì F(x) chia hết cho x + 1
\(\left(x+3\right)^3-x\left(3x+1\right)^2+\left(2x+1\right)\left(4x^2-2x+1\right)=28\)
\(\Leftrightarrow x^3+9x^2+27x+27-9x^3-6x^2-x+8x^3+1=28\)
\(\Leftrightarrow3x^2+26x+28=28\)
\(\Leftrightarrow3x^2+26x=0\)\(\Leftrightarrow x\left(3x+26\right)=0\)
Suy ra x=0 hoặc x=-26/3
cho mk hỏi ngu tí là 6x^2 ở đâu thế ạ