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\(5x^2\left(3x-2\right)-3x^2\left(5x+2\right)+2x\left(3+8x\right)=21\)
\(\Leftrightarrow15x^3-10x^2-15x^3-6x^2+6x+16x^2-21=0\)
\(\Leftrightarrow6x-21=0\)
\(\Leftrightarrow6x=21\)
\(\Leftrightarrow x=3,5\)
a: \(N=\dfrac{3x^5-4x^4+6x^3}{-2x^2}=-\dfrac{3}{2}x^3+2x^2-3x\)
b: \(N=\dfrac{\left(6x^4y^5-3x^3y^4+\dfrac{1}{2}x^4y^3z\right)}{-\dfrac{1}{3}x^2y^3}=-18x^2y^2+9xy-\dfrac{3}{2}x^2z\)
c: \(\Leftrightarrow N\cdot\left(y-x\right)=\left(x-y\right)^3\)
\(\Leftrightarrow N=\dfrac{\left(x-y\right)^3}{y-x}=-\left(y-x\right)^2\)
d: \(\Leftrightarrow N\cdot\left(y^2-x^2\right)=\left(y^2-x^2\right)^2\)
hay \(N=y^2-x^2\)
a) \(ĐKXĐ:\hept{\begin{cases}x\ne\pm2\\x\ne-3\end{cases}}\)
b) \(P=1+\frac{x+3}{x^2+5x+6}\div\left(\frac{8x^2}{4x^3-8x^2}-\frac{3x}{3x^2-12}-\frac{1}{x+2}\right)\)
\(\Leftrightarrow P=1+\frac{x+3}{\left(x+3\right)\left(x+2\right)}:\left(\frac{8x^2}{4x^2\left(x-2\right)}-\frac{3x}{3\left(x^2-4\right)}-\frac{1}{x+2}\right)\)
\(\Leftrightarrow P=1+\frac{1}{x+2}:\left(\frac{2}{x-2}-\frac{x}{\left(x-2\right)\left(x+2\right)}-\frac{1}{x+2}\right)\)
\(\Leftrightarrow P=1+\frac{1}{x+2}:\frac{2x+4-x-x+2}{\left(x-2\right)\left(x+2\right)}\)
\(\Leftrightarrow P=1+\frac{1}{x+2}:\frac{6}{\left(x-2\right)\left(x+2\right)}\)
\(\Leftrightarrow P=1+\frac{\left(x-2\right)\left(x+2\right)}{6\left(x+2\right)}\)
\(\Leftrightarrow P=1+\frac{x-2}{6}\)
\(\Leftrightarrow P=\frac{x+4}{6}\)
c) Để P = 0
\(\Leftrightarrow\frac{x+4}{6}=0\)
\(\Leftrightarrow x+4=0\)
\(\Leftrightarrow x=-4\)
Để P = 1
\(\Leftrightarrow\frac{x+4}{6}=1\)
\(\Leftrightarrow x+4=6\)
\(\Leftrightarrow x=2\)
d) Để P > 0
\(\Leftrightarrow\frac{x+4}{6}>0\)
\(\Leftrightarrow x+4>0\)(Vì 6>0)
\(\Leftrightarrow x>-4\)
a: \(=\left(x+1+5\right)\left(x+1-5\right)=\left(x+6\right)\left(x-4\right)\)
b: =(1-2x)(1+2x)
c: \(=\left(2-3x\right)\left(4+6x+9x^2\right)\)
d: =(x+3)^3
e: \(=\left(2x-y\right)^3\)
f: =(x+2y)(x^2-2xy+4y^2)
a) 4x(3x-7)-6(2x2-5x+1)=12
=>4x.3x-4x.7-6.2x2-6.(-5x)-6.1=12
=>12x2-28x-12x2+30x-6=12
=>2x-6 =12
=>2x =12+6
=>2x =18
=>x =18:2
=>x =6
b)(5x+3)(4x-1)+(10x-7)(-2x+3)=27
=>5x.4x-5x.1+3.4x+3.(-1)+10x.(-2x)+10x.3-7.-(2x)-7.3=27
=>20x2-5x+12x-3-20x2+30x+14x-21=27
=>39x-36 =27
=>39x =27+36
=>39x =63
=>x =63:39
=>x =21/13
c) (8x-5)(3x+2)-(12x+7)(2x-1)=17
=>8x.3x+8x.2-5.3x-5.2-12x.2x-12x.(-1)+7.2x+7.(-1)=17
=>24x2+16x-15x-10-24x2+12x+14x-7=17
=>27x-17 =17
=>27x =17+17
=>27x =34
=>x =34:27
=>x =34/27
d) (5x+9)(6x-1)-(2x-3)(15x+1)=-190
=>30x2-5x+63x-9 - 30x2-2x-45x-3=-190
=>11x-12 =-190
=>11x =-190+12
=>11x =-178
=>x = -178:11
=>x =-178/11
a: ta có: \(\left(8x+2\right)\left(1-3x\right)+\left(6x-1\right)\left(4x-10\right)=-50\)
\(\Leftrightarrow8x-24x^2+2-6x+24x^2-60x-4x+40=-50\)
\(\Leftrightarrow-62x=-92\)
hay \(x=\dfrac{46}{31}\)
b: ta có: \(\left(1-4x\right)\left(x-1\right)+4\left(3x+2\right)\left(x+3\right)=38\)
\(\Leftrightarrow x-1-4x^2+4x+4\left(3x^2+9x+2x+6\right)=38\)
\(\Leftrightarrow-4x^2+5x-1+12x^2+44x+24-38=0\)
\(\Leftrightarrow8x^2+49x-15=0\)
\(\text{Δ}=49^2-4\cdot8\cdot\left(-15\right)=2881\)
Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{-49-\sqrt{2881}}{16}\\x_2=\dfrac{-49+\sqrt{2881}}{16}\end{matrix}\right.\)
(3x-5)^2 - (3x+1)^2 =8
(3x)^2 - 2*3x*5 + 5^2 -[(3x)^2 + 2*3x*5 + 1^2]= 8
9x^2 - 30x + 25 - (9x^2 + 30x + 1) = 8
9x^2 - 30x + 25 - 9x^2 - 30x - 1 = 8
- 30x + 25 - 30x - 1 = 8
2*(-30x) + (25 - 1) = 8
-60x + 24 = 8
-60x = 8 - 24
-60x = -16
x = -16 / -60
x = 16 / 60
x = 16 * 1/60
x = 16/60
x = 4/15
mình giải chi tiết bạn tự rút gọn nhé sợ bạn ko hiểu