\(\dfrac{y+z}{x}+\dfrac{x+z}{y}+\dfrac{x+y}{z}+3\)

\(< =>\dfrac{y+z}{x}+1+\dfrac{x+z}{y}+1+\dfrac{x+y}{z}+1\)

\(< =>\dfrac{y+z}{x}+\dfrac{x}{x}+\dfrac{x+z}{y}+\dfrac{y}{y}+\dfrac{x+y}{z}+\dfrac{z}{z}\)

\(< =>\dfrac{x+y+z}{x}+\dfrac{x+y+z}{y}+\dfrac{x+y+z}{z}\) (1)

Thay x+y+z=0vào (1), ta có:\(\dfrac{0}{x}+\dfrac{0}{y}+\dfrac{0}{z}=0+0+0=0\)

Số 18 là bội của 3, vì 18:3=6

Số 18 không phải là bội của 4, vì 18:4=4 dư 2

Số 4 là ước của 12,vì 12:4=3 hay 3.4=12

Số 4 không phải là ước của 15, vì 15:4=3 dư 3

\(\dfrac{cos}{1-sin}=\dfrac{1+sin}{cos}=>cos.cos=\left(1-sin\right)\left(1+sin\right)\)

\(=>cos^2=1-sin^2=>cos^2+sin^2=1\)

\(=>\dfrac{k^2}{h^2}+\dfrac{đ^2}{h^2}=1=>\dfrac{k^2+đ^2}{h^2}=1\)

\(=>k^2+đ^2=h^2(ĐLPY-TA-GO)\)

\(\dfrac{\left(sin+cos\right)^2-\left(sin-cos\right)^2}{sin.cos}\)

\(=\dfrac{\left(sin+cos-sin+cos\right)\left(sin+cos+sin-cos\right)}{sin.cos}\)

\(=\dfrac{2sin.2cos}{sin.cos}=\dfrac{4sin.cos}{sin.cos}=4\)

a)\(\sqrt{\dfrac{4x+3}{x+1}}=3< =>\dfrac{4x+3}{x+1}=3^2=9\)

\(=>4x+3=9\left(x+1\right)=9x+9\)

\(=>4x-9x=9-3< =>-5x=6\)

\(=>x=\dfrac{-6}{5}\)

b)\(\dfrac{1}{2}\sqrt{4x-8}-\dfrac{2}{3}\sqrt{x-2}+\sqrt{\dfrac{x-2}{36}}=7\)

\(< =>\dfrac{1}{2}\sqrt{2^2\left(x-2\right)}-\dfrac{2}{3}\sqrt{x-2}+\sqrt{\dfrac{x-2}{6^2}}=7\)

\(< =>\dfrac{1}{2}.2\sqrt{x-2}-\dfrac{2}{3}\sqrt{x-2}+\dfrac{1}{6}\sqrt{x-2}=7\)

\(< =>\left(1-\dfrac{2}{3}+\dfrac{1}{6}\right)\sqrt{x-2}=7\)

\(< =>\dfrac{1}{2}\sqrt{x-2}=7< =>\sqrt{x-2}=7:\dfrac{1}{2}=14\)

\(< =>x-2=14^2=196< =>x=196+2=198\)

(x-3)(3x-9)=0

<=>(x-3)3(x-3)=0

=>x-3=0=>x=3

U=R₁.I₁=4R₂.I₁

U=R₂.I₂=R₂.(I₁+6)

=>4R₂.I₁=R₂(I₁+6)

=>4I₁=I₁+6=>3I₁=6=>I₁=2(A)

I₂=I₁+6=2+6=8(A)

R₁=U/I₁=16/2=8(ôm)

R₂=U/I₂=16/8=2(ôm)

(a+b+c)²=a²+2ab+b²+2bc+c²+2ac

a) x(x+5)

b) 2x+1

c) x-2

d) x+2

1001=abcabc:abc

=>abc*1001=abc.abcabc:abc=abcabc

ab*1001=ab.abcabc:abc=ababc