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a) \(5xy\cdot\left(-2bx^2y\right)=-10b\left(x\cdot x^2\right)\left(y\cdot y\right)=-10bx^3y^2\)
b) \(\left(-\frac{4}{5}ab^2c\right)\left(-20a^4bx\right)=\left[\left(-\frac{4}{5}\right)\cdot\left(-20\right)\right]\left(a\cdot a^4\right)\left(b^2\cdot b\right)cx\)
\(=16a^5b^3cx\)
c) \(2^3abc\cdot\frac{1}{4}a^2bc^3=8abc\cdot\frac{1}{4}a^2bc^3=2\left(a\cdot a^2\right)\left(b\cdot b\right)\left(c\cdot c^3\right)=2a^3b^2c^4\)
d) \(a^3b^3a^2b^2c=\left(a^3\cdot a^2\right)\left(b^3\cdot b^2\right)c=a^5b^5c\)
e) \(2ab\cdot\frac{4}{3}a^2b^4\cdot7abc=\left(2\cdot\frac{4}{3}\cdot7\right)\left(a\cdot a^2\cdot a\right)\left(b\cdot b^4\cdot b\right)c=\frac{56}{3}a^4b^6c\)
f) \(\left(-1,5ab^2\right)\cdot\frac{1}{4}bca^2b=\left(-1,5\cdot\frac{1}{4}\right)\left(a\cdot a^2\right)\left(b^2\cdot b\cdot b\right)c=-\frac{3}{8}a^3b^4c\)
a) \(5xy.\left(-2bx^2y\right)\)
\(=\left[5.\left(-2\right)\right]\left(x.x^2\right)\left(y.y\right).b\)
\(=-10x^3y^2b\)
b) \(\left(-\frac{4}{5}ab^2c\right)\left(-20a^4bx\right)\)
\(=\left[\left(-\frac{4}{5}\right)\left(-20\right)\right]\left(a.a^{4\:}\right)\left(b^2b\right).c.x\)
\(=16a^5b^3cx\)
c) \(2^3abc.\frac{1}{4}a^2bc^3\)
\(=\left(2^3.\frac{1}{4}\right)\left(aa^{2\:}\right)\left(bb\right)\left(cc^3\right)\)
\(=2a^3b^2c^4\)
d) \(a^3b^3a^2b^2c\)
\(=\left(a^3a^2\right)\left(b^3b^2\right)c\)
\(=a^5b^5c\)
e) \(2ab.\frac{4}{3}a^2b^47abc\)
\(=\left(2.\frac{4}{3}.7\right)\left(aa^{2\: }a\right)\left(bb^4b\right)c\)
\(=\frac{56}{3}a^4b^6c\)
f) \(\left(-1,5ab^2\right)\frac{1}{4}bca^2b\)
\(=\left(-1,5.\frac{1}{4}\right)\left(aa^{2\:}\right)\left(b^2bb\right)\)
\(=-\frac{3}{8}a^3b^4\)
Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow\hept{\begin{cases}a=bk\\c=dk\end{cases}}\)
Khi đó, ta có : \(\frac{3bk+2b}{2bk+3b}=\frac{\left(3k+2\right)b}{\left(2k+3\right)b}=\frac{3k+2}{2k+3}\)(1)
\(\frac{3dk+2d}{2dk+3d}=\frac{\left(3k+2\right).d}{\left(2k+3\right).d}=\frac{3k+2}{2k+3}\)(2)
Từ (1) và (2), suy ra : \(\frac{3a+2b}{2a+3b}=\frac{3c+2d}{2c+3d}\)
Đặt ab = x, bc = y, ca = z (x, y, z ≠ 0 thỏa mãn x^3 + y^3 + z^3 = 3xyz)
⇔ (x+y)^3 − 3xy(x + y) + z^3 = 3xyz <=> (x+y)^3 − 3xy(x + y) + z^3 = 3xyz
⇔ (x + y)^3 + z^3 − 3xy(x + y+ z) = 0 ⇔ (x + y)^3 + z^3 − 3xy(x + y + z) = 0
⇔ (x + y + z)[(x + y)^2 − z (x + y) + z^2] − 3xy(x + y + z) = 0 ⇔ (x + y + z)[(x + y)^2 − z(x + y) + z2] − 3xy(x + y + z) = 0
⇔ (x + y + z)(x^2 + y^2 + z^2 − xy − yz − xz) = 0 ⇔ (x + y + z)(x^2 + y^2 + z^2 − xy − yz − xz) = 0
<=> x + y + z = 0 (1) và x^2 + y^2 + z^2 − xy − yz − xz = 0 (2)
Với (1): ⇔ ab + bc + ac = 0 ⇔ ab + bc + ac = 0
P = (1 + a/b)(1 + b/c)(1 + c/a) = (a + b)(b + c)(c + a)/abc=(ab + bc + ac)(a + b + c) − abc/abc = 0 − abc/abc = −1
Với (2) ⇔ (x − y)^2 + (y − z)^2 + (z − x)^2/2 = 0
⇔ (x − y)^2 + (y − z)^2 + (z − x)^2 = 0
Ta thấy (x − y)^2; (y − z)^2; (z − x)^2 ≥ 0 ∀x, y, z nên để tổng của chúng bằng 0 thì:
(x − y)^2 = (y − z)^2 = (z − x)^2 = 0 ⇒ x = y = z
⇔ ab = bc = ac ⇔ a=b=c (do a, b, c ≠ 0)
⇒ A = (1 + 1)(1 + 1)(1 + 1) = 8
Vậy...........
Đặt a/b=c/d=k
=>a=bk; c=dk
a: \(\dfrac{3a-c}{3b-d}=\dfrac{3bk-dk}{3b-d}=k\)
\(\dfrac{2a+3c}{2b+3d}=\dfrac{2bk+3dk}{2b+3d}=k\)
Do đó: \(\dfrac{3a-c}{3b-d}=\dfrac{2a+3c}{2b+3d}\)
c: \(\dfrac{a^2-b^2}{c^2-d^2}=\dfrac{b^2k^2-b^2}{d^2k^2-d^2}=\dfrac{b^2}{d^2}\)
\(\dfrac{2ab+b^2}{2cd+d^2}=\dfrac{2\cdot bk\cdot b+b^2}{2\cdot dk\cdot d+d^2}=\dfrac{b^2}{d^2}\)
Do đó: \(\dfrac{a^2-b^2}{c^2-d^2}=\dfrac{2ab+b^2}{2cd+d^2}\)
a: \(=ab\cdot\dfrac{4}{3}a^2b^4\cdot7abc=\dfrac{28}{3}a^4b^6c\)
b: \(a^3b^3\cdot a^2b^2c=a^5b^5c\)
c: \(=\dfrac{2}{3}a^3b\cdot\dfrac{-1}{2}ab\cdot a^2b=\dfrac{-1}{3}a^6b^3\)
d: \(=-\dfrac{7}{3}a^3c^2\cdot\dfrac{1}{7}ac^2\cdot6abc=-2a^5bc^5\)
e: \(=\dfrac{-3}{2}\cdot\dfrac{1}{4}\cdot ab^2\cdot bca^2\cdot b=\dfrac{-3}{8}a^3b^4c\)