A ^ B = 8
B ^ A = 6
A - B = -1
B - A = 1
A + B = 5
B + A = 5
B / A = 1,5
tính C biết C bằng A ^ B x A ^ B x B ^ A x B ^ A
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a) \(\left(2a-b\right)\left(b+4a\right)+2a\left(b-3a\right)\)
\(=2ab+8a^2-b^2-4ab+2ab-6a^2\)
\(=\left(2ab+2ab-4ab\right)+\left(8a^2-6a^2\right)-b^2\)
\(=2a^2-b^2\)
b) \(\left(3a-2b\right).\left(2a-3b\right)-6a\left(a-b\right)\)
\(=6a^2-9ab-4ab+6b^2-6a^2+6ab\)
\(=\left(6a^2-6a^2\right)-\left(9ab+4ab-6ab\right)+6b^2\)
\(=-7ab+b^2\)
c) \(5b\left(2x-b\right)-\left(8b-x\right)\left(2x-b\right)\)
\(=10bx-5b^2-\left(16bx-8b^2-2x^2+bx\right)\)
\(=10bx-5b^2-16bx+8b^2+2x^2-bx\)
\(=\left(10bx-16bx-bx\right)-\left(5b^2-8b^2\right)+2x^2\)
\(=-7bx+3b^2+2x^2\)
d) \(2x\left(a+15x\right)+\left(x-6a\right)\left(5a+2x\right)\)
\(=2ax+30x^2+5ax+2x^2-30a^2-12ax\)
\(=\left(2ax+5ax-12ax\right)+\left(30x^2+2x^2\right)-30a^2\)
\(=-5ax+32x^2-30a^2\)
a: =2ab+8a^2-b^2-4ab+2ab-6a^2
=2a^2-b^2
b: =6a^2-9ab-4ab+6b^2-6a^2+6ab
=-7ab+6b^2
c: =10bx-5b^2-16bx+8b^2+2x^2-xb
=3b^2+2x^2-7xb
d: =2xa+30x^2+5ax+2x^2-30a^2-12ax
=32x^2-30a^2-5ax
a) (2a - b)(b + 4a) + 2a(b - 3a)
= 2a(b + 4a) - b(b + 4a) + 2ab - 6a^2
= 2ab + 8a^2 - b^2 - 4ab + 2ab - 6a^2
= (8a^2 - 6a^2) + (2ab + 2ab - 4ab) - b^2
= 2a^2 - b^2
b) .(3a - 2b)(2a - 3b) - 6a(a - b)
= 3a(2a - 3b) - 2b(2a - 3b) - (6a^2 - 6ab)
= 6a^2 - 9ab - (4ab - 6b^2) - (6a^2 - 6ab)
= 6a^2 - 9ab - 4ab + 6b^2 - 6a^2 + 6ab
= 6b^2 + (6a^2 - 6a^2) + (6ab - 4ab - 9ab)
= 6b^2 - 7ab
c. 5b(2x - b) - (8b - x)(2x - b)
= 10bx - 5b^2 - 8b(2x - b) + x(2x - b)
= 10bx - 5b^2 - 16bx + 8b^2 + 2x^2 - bx
= (10bx - 16bx - bx) + 2x^2 + (8b^2 - 5b^2)
= -7bx + 2x^2 + 3b^2
d. 2x(a + 15x) + (x - 6a)(5a + 2x)
= 2ax + 30x^2 + x(5a + 2x) - 6a(5a + 2x)
= 2ax + 30x^2 + 5ax + 2x^2 - 30a^2 - 12ax
= (30x^2 + 2x^2) + (2ax + 5ax - 12ax) - 30a^2
= 32x^2 - 5ax - 30a^2
Chúc bạn hok tốt !!!
a: 3x=2y
nên x/2=y/3
Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{x-y}{2-3}=\dfrac{1}{-1}=-1\)
Do đó: x=-2; y=-3
\(A=\left(-2\right)^3+12\cdot\left(-2\right)^2\cdot\left(-3\right)+48\cdot\left(-2\right)\cdot\left(-3\right)^2-64\cdot\left(-3\right)^3\)
\(=-8+12\cdot4\cdot\left(-3\right)-96\cdot9-64\cdot\left(-27\right)\)
\(=712\)
b: 6a=5b
nên a/5=b/6
Đặt a/5=b/6=k
=>a=5k; b=6k
\(B=\dfrac{2a-3b}{3b-2a}=-1\)
d: \(\left|x-2\right|+\left(y-1\right)^2=0\)
=>x-2=0 và y-1=0
=>x=2 và y=1
\(D=\left|2-2\right|+\dfrac{2-1}{2-1}=0+1=1\)
a: \(P\left(x\right)=6x^3-x-1+5x^3+x^2-2x-1=11x^3+x^2-3x-2\)
b: \(Q\left(x\right)=6x^3-x-1-5x^3-x^2+2x+1=x^3-x^2+x\)
c: \(2\cdot U\left(x\right)=5x^3-2x+x^2-1-3\cdot\left(6x^3-x-1\right)\)
\(\Leftrightarrow2\cdot U\left(x\right)=5x^3+x^2-2x-1-18x^3+3x+3\)
\(\Leftrightarrow2\cdot U\left(x\right)=-13x^3+x^2+x+2\)
hay \(U\left(x\right)=-\dfrac{13}{2}x^3+\dfrac{1}{2}x^2+\dfrac{1}{2}x+1\)
\(a,P\left(x\right)=A\left(x\right)+B\left(x\right)=6x^3-x-1+5x^3-2x+x^2-1=11x^3+x^2+x-2\)
\(b,Q\left(x\right)=A\left(x\right)-B\left(x\right)=6x^3-x-1-5x^3+2x-x^2+1=x^3-x^2+x\)
c, Ta có : \(2U\left(x\right)+2A\left(x\right)=B\left(x\right)\)
hay \(2U\left(x\right)+2\left(6x^3-x-1\right)=5x^3-2x+x^2-1\)
\(\Rightarrow2U\left(x\right)+12x^3-2x-2=5x^3-2x+x^2-1\)
\(\Rightarrow2U\left(x\right)=5x^3-2x+x^2-1-12x^3+2x+2\)
\(\Rightarrow U\left(x\right)=-7x^3+x^2+1\)
a) Nếu a = 4, b = 3 và c = 5 thì a x b x c = 4 x 3 x 5 = 12 x 5 = 60
b) Nếu a = 21, b = 0, c = 58 thì a x b x c = 21 x 0 x 58 = 0 x 58 = 0
\(\left(7x-4\right)\left(2x+3\right)-13x\)
\(=14x^2+21x-8x-12-13x\)
\(=14x^2-12\)
\(a^3-\left(a^2-3a\right)\left(a+3\right)\)
\(=a^3-\left(a^3+3a^2-3a^2-9a\right)\)
\(=a^3-a^3-3a^2+3a^2+9a\)
\(=9a\)
\(\left(2a-b\right)\left(b+4a\right)+2a\left(b-3a\right)\)
\(=2ab+8a^2-b^2-4ab+2ab-6a^2\)
\(=\)\(2a^2-b^2\)
\(5b\left(2x-b\right)+\left(x-6a\right)\left(5a+2x\right)\)
\(=10bx-5b^2+5ax+2x^2-30a^2-12ax\)
\(=2x^2-30a^2-5b^2+10bx-7ax\)
a) Tìm x
\(6-\left(x-\frac{1}{3}\right)^2=\frac{2^{2013}}{\left(-2\right)^{2012}}\Rightarrow6-\left(x-\frac{1}{3}\right)^2=\frac{2^{2013}}{2^{2012}}=2^1=2\)
\(\Rightarrow\left(x-\frac{1}{3}\right)^2=6-2=4=2^2\Rightarrow\hept{\begin{cases}x-\frac{1}{3}=2\\x-\frac{1}{3}=-2\end{cases}\Rightarrow\hept{\begin{cases}x=\frac{7}{3}\\x=\frac{-5}{3}\end{cases}}}\)
Vậy \(x\in\left\{\frac{7}{3};\frac{-5}{3}\right\}\)
b) Ta có : \(2a=3b\Rightarrow\frac{a}{3}=\frac{b}{2}\) và \(5b=7c\Rightarrow\frac{b}{7}=\frac{c}{5}\)
\(\Rightarrow\hept{\begin{cases}\frac{a}{3}=\frac{b}{2}\Rightarrow\frac{a}{21}=\frac{b}{14}\\\frac{b}{7}=\frac{c}{5}\Rightarrow\frac{b}{14}=\frac{c}{10}\end{cases}}\Rightarrow\frac{a}{21}=\frac{b}{14}=\frac{c}{10}\)
Áp dụng tính chất dãy tỉ số bằng nhau, ta có : \(\frac{a}{21}=\frac{b}{14}=\frac{c}{10}=\frac{a+b-c}{21+14-10}=-\frac{50}{25}=-2\)
\(\Rightarrow a=\left(-2\right).21=-42\) \(b=\left(-2\right).14=-28\) \(c=\left(-2\right).5=-10\)
Vậy a = -42 ; b = -28 và c = -10
Từ A+B =5 => B= 5-A
Thay B= 5-A Vào B-A = 1 . Ta được :
5-A-A=1 <=> -2a=-4 <=> A= 2
THAY A=2 Vào A+B = 5 . Được : 2+B=5 => B=3
Vậy C= 23.23.32.32=8.8.9.9=5184