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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) \(\left(-1,5ab^2\right)\frac{1}{4}bca^2b\)
\(=\left(-1,5.\frac{1}{4}\right)\left(aa^{2\:}\right)\left(b^2bb\right)\)
\(=-\frac{3}{8}a^3b^4\)
a, a/b=c/d
<=>a/c=b/d
<=>2a/2c=3b/3d=2a+3b/2c+3d=2a-3b/2c-3d
<=>2a+3b/2a-3b=2c+3d/2c-3d(đpcm)
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) \(\left(-1,5ab^2\right)\cdot\frac{1}{4}bca^2b=\left(-1,5\cdot\frac{1}{4}\right)\left(a\cdot a^2\right)\left(b^2\cdot b\cdot b\right)c=-\frac{3}{8}a^3b^4c\)
Lời giải:
a.
$f(-1)=a-b+c$
$f(-4)=16a-4b+c$
$\Rightarrow f(-4)-6f(-1)=16a-4b+c-6(a-b+c)=10a+2b-5c=0$
$\Rightarrow f(-4)=6f(-1)$
$\Rightarrow f(-1)f(-4)=f(-1).6f(-1)=6[f(-1)]^2\geq 0$ (đpcm)
b.
$f(-2)=4a-2b+c$
$f(3)=9a+3b+c$
$\Rightarrow f(-2)+f(3)=13a+b+2c=0$
$\Rightarrow f(-2)=-f(3)$
$\Rightarrow f(-2)f(3)=-[f(3)]^2\leq 0$ (đpcm)
a.
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Áp dụng tính chất dãy tỉ số bằng nhau ta có :
\(\frac{2b+c-a}{a}=\frac{2c-b+a}{b}=\frac{2a+b-c}{c}=\frac{2a+2b+2c}{a+b+c}=\frac{2\left(a+b+c\right)}{a+b+c}=2\)
Do đó :
\(\frac{2b+c-a}{a}=2\)\(\Rightarrow\)\(c=3a-2b\)\(;\)\(2b=3a-c\)\(\left(1\right)\)
\(\frac{2c-b+a}{b}=2\)\(\Rightarrow\)\(a=3b-2c\)\(;\)\(2c=3b-a\)\(\left(2\right)\)
\(\frac{2a+b-c}{c}=2\)\(\Rightarrow\)\(b=3c-2a\)\(;\)\(2a=3c-b\)\(\left(3\right)\)
Thay (1), (2) và (3) vào \(P=\frac{\left(3a-2b\right)\left(3b-2c\right)\left(3c-2a\right)}{\left(3a-c\right)\left(3b-a\right)\left(3c-b\right)}\) ta được :
\(P=\frac{c.a.b}{2b.2c.2a}=\frac{abc}{8abc}=\frac{1}{8}\)
Vậy \(P=\frac{1}{8}\)
Chúc bạn học tốt ~
Phùng Minh Quân sai nha nếu a+b+c = 0 thì a+b+c / 2(a+b+c) thì nó không bằng 1/2 đc mà nó bằng 0
a/ Ta có: \(b^2=ac\Rightarrow\frac{a}{b}=\frac{b}{c};c^2=bd\Rightarrow\frac{b}{c}=\frac{c}{d}\)\(\Rightarrow\frac{a}{b}=\frac{b}{c}=\frac{c}{d}\)
Đặt \(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=k\Rightarrow\left(\frac{a}{b}\right)^3=\left(\frac{b}{c}\right)^3=\left(\frac{c}{d}\right)^3=k^3\Rightarrow\frac{a^3}{b^3}=\frac{b^3}{c^3}=\frac{c^3}{d^3}=k^3\)
Áp dụng tính chất của tỉ lệ thức ta có:\(\frac{a^3}{b^3}=\frac{b^3}{c^3}=\frac{c^3}{d^3}=\frac{a^3+b^3+c^3}{b^3+c^3+d^3}=k^3\)
Mặt khác: \(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=k\Rightarrow\frac{a+b+c}{b+c+d}=k\Rightarrow\left(\frac{a+b+c}{b+c+d}\right)^3=k^3\)
\(\Rightarrow\frac{a^3+b^3+c^3}{b^3+c^3+d^3}=\left(\frac{a+b+c}{b+c+d}\right)^3\left(=k^3\right)\)
