c: \(-\dfrac{1}{5}:\left(1\dfrac{2}{5}\right)=-\dfrac{2}{7}:x\)

=>\(-\dfrac{2}{7}:x=-\dfrac{1}{5}:\dfrac{7}{5}=\dfrac{-1}{5}\cdot\dfrac{5}{7}=-\dfrac{1}{7}\)

=>\(x=\dfrac{-2}{7}:\dfrac{-1}{7}=\dfrac{2}{7}\cdot\dfrac{7}{1}=2\)

d: \(1\dfrac{1}{2}:\dfrac{2}{3}=x:\dfrac{1}{9}\)

=>\(x:\dfrac{1}{9}=\dfrac{3}{2}\cdot\dfrac{3}{2}=\dfrac{9}{4}\)

=>\(x=\dfrac{9}{4}\cdot\dfrac{1}{9}=\dfrac{1}{4}\)

Em cảm ơn 

a: \(\dfrac{2}{5}+\dfrac{3}{5}:\left(-\dfrac{3}{2}\right)+\dfrac{1}{2}\)

\(=\dfrac{2}{5}+\dfrac{3}{5}\cdot\dfrac{-2}{3}+\dfrac{1}{2}\)

\(=\dfrac{2}{5}-\dfrac{2}{5}+\dfrac{1}{2}=\dfrac{1}{2}\)

b: \(2,5-\left(-\dfrac{5}{6}\right)^0+\left(-\dfrac{1}{6}\right)^2\cdot\left(-3\right)\)

\(=\dfrac{5}{2}-1+\dfrac{1}{36}\cdot\left(-3\right)\)

\(=\dfrac{3}{2}-\dfrac{1}{12}=\dfrac{18}{12}-\dfrac{1}{12}=\dfrac{17}{12}\)

Lời giải:
Có:
$(a^2+1)(b^2+1)(c^2+1)=(a^2+ab+bc+ac)(b^2+ab+bc+ac)(c^2+ab+bc+ac)$

$=(a+b)(a+c)(b+c)(b+a)(c+a)(c+b)=[(a+b)(b+c)(c+a)]^2$

Và:

$(a+b+c-abc)^2=[(a+b+c)(ab+bc+ac)-abc]^2$

$=[ab(a+b)+bc(b+c)+ca(c+a)+2abc]^2$

$=[ab(a+b+c)+bc(b+c+a)+ca(c+a)]^2$

$=[(a+b+c)(ab+bc)+ca(c+a)]^2=[b(a+b+c)(a+c)+ac(c+a)]^2$

$=[(c+a)(ab+b^2+bc+ac)]^2=[(c+a)(b+a)(b+c)]^2$
Do đó: $P=\frac{[(a+b)(b+c)(c+a)]^2}{[(a+b)(b+c)(c+a)]^2}=1$

LÀM ƠN GIÚP MIK ĐI MÀ, NĂN NỈ CÁC BẠN ĐÓ!!!!!!!!!!!!!!!!!!!!!!

Lời giải:

Với $ab+bc+ac=1$ ta có:

$a^2+1=a^2+ab+bc+ac=(a+b)(a+c)$

$b^2+1=b^2+ab+bc+ac=(b+c)(b+a)$

$c^2+1=c^2+ab+bc+ac=(c+a)(c+b)$

$\Rightarrow (a^2+1)(b^2+1)(c^2+1)=[(a+b)(b+c)(c+a)]^2(*)$

Mặt khác:

$a+b+c-abc=(a+b+c)(ab+bc+ac)-abc$

$=ab(a+b)+bc(b+c)+ca(c+a)+2abc$

$=ab(a+b+c)+bc(b+c)+ca(c+a)$
$=(a+b+c)(ab+bc)+ca(c+a)=b(a+b+c)(c+a)+ca(c+a)$

$=(c+a)[b(a+b+c)+ca]=(c+a)(b+a)(b+c)$

$\Rightarrow (a+b+c-abc)^2=[(a+b)(b+c)(c+a)]^2(**)$

Từ $(*); (**)\Rightarrow P=1$

16.79

16,79

\(0,3:2,5=3:25\)

\(4\dfrac{2}{5}:1\dfrac{1}{3}=\dfrac{22}{5}:\dfrac{4}{3}=33:10\)

\(-3,2:1\dfrac{2}{7}=\dfrac{-16}{5}:\dfrac{9}{7}=112:45\)

Lời giải:

a. $\frac{3}{-2}=\frac{9}{-6}$

$\frac{-2}{3}=\frac{-6}{9}$

$\frac{3}{9}=\frac{-2}{-6}$

$\frac{9}{3}=\frac{-6}{-2}$

b.

$\frac{12}{16}=\frac{3}{4}$

$\frac{16}{12}=\frac{4}{3}$

$\frac{12}{3}=\frac{16}{4}$

$\frac{3}{12}=\frac{4}{16}$