Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow a=bk;c=dk\)

\(VT=\left(\dfrac{a+b}{c+d}\right)^2=\left(\dfrac{kb+b}{dk+d}\right)^2=\left(\dfrac{b\left(k+1\right)}{d.\left(k+1\right)}\right)^2=\dfrac{b^2}{d^2}\)

\(VP=\dfrac{a^2-b^2}{c^2-d^2}=\dfrac{b^2k^2-b^2}{d^2k^2-d^2}=\dfrac{b^2\left(k^2-1\right)}{d^2\left(k^2-1\right)}=\dfrac{b^2}{d^2}\)

VT=VP => dpcm =))

=)) quỳ

-đưa sang dạng xâu rồi chạy for tìm được cuối với đầu

-cách 2 thì: lấy nó chia dư cho 10 thì thu dc số cuối , chia nguyên nó cho 10 cho đến khi nó nhỏ hơn 10 thì dừng :v

éo chắc

ĐK: x >-3/2 và y khác y\(\ge\)0

\(\dfrac{y}{2x+3}=\dfrac{\sqrt{2x+3}+1}{\sqrt{y}+1}\)

=>\(\left(\sqrt{y}\right)^3+\left(\sqrt{y}\right)^2=\left(\sqrt{2x+3}\right)^3+\left(\sqrt{2x+3}\right)^2\)

<=>\(\left(\sqrt{y}-\sqrt{2x+3}\right)\left(2x+3+y+\sqrt{y}\sqrt{2x+3}\right)+\left(\sqrt{y}-\sqrt{2x+3}\right)\left(\sqrt{y}+\sqrt{2x+3}\right)=0\)

<=>\(\left(\sqrt{y}-\sqrt{2x+3}\right)\left(2x+3+y+\sqrt{y}\sqrt{2x+3}+\sqrt{y}+\sqrt{2x+3}\right)=0\)

<=>\(\sqrt{y}=\sqrt{2x+3}\)(\(2x+3+y+\sqrt{y}\sqrt{2x+3}+\sqrt{y}+\sqrt{2x+3}\ne0\))

<=>y=2x+3

Suy ra: Q=2x²+3x-6x-9-2x-3

=2x²-5x-12

=2(x²-2.x.\(\dfrac{5}{4}\)+\(\dfrac{25}{16}\)-\(\dfrac{121}{16}\))

=2(x-\(\dfrac{5}{4}\))²-\(\dfrac{121}{8}\)\(\ge\dfrac{-121}{8}\)

Dấu "=" xảy ra khi x=5/4 =>y=11/2

Xấu ***** chắc sai

Chỉ cần STN\(\notin\)B(30) thì không chia hết cho 30 và có dư

yêu cầu là gì

ĐK: \(x\ne k\pi;x\ne\dfrac{\pi}{2}+k\pi\left(k\in Z\right)\)

\(8cos2x=\dfrac{\sqrt{3}}{sinx}+\dfrac{1}{cosx}\Rightarrow8cos2x.sinx.cosx=\sqrt{3}cosx+sinx\)

<=>4cos2x.sin2x=\(\sqrt{3}\)cosx+sinx

<=>2cos4x=\(\sqrt{3}\)cosx+sinx

<=>cos4x=\(\dfrac{\sqrt{3}}{2}cosx+\dfrac{1}{2}sinx\)

<=>cos4x=cos\(\dfrac{\pi}{6}\).cosx+sin\(\dfrac{\pi}{6}\).sinx

<=>cos4x=cos(\(\dfrac{\pi}{6}\)-x)

<=>\(\left[{}\begin{matrix}4x=\dfrac{\pi}{6}-x+k2\pi\\4x=\pi-\left(\dfrac{\pi}{6}-x\right)+k2\pi\end{matrix}\right.\)

<=>\(\left[{}\begin{matrix}x=\dfrac{\pi}{30}+k\dfrac{2\pi}{5}\\x=\dfrac{5\pi}{18}+k\dfrac{2\pi}{3}\end{matrix}\right.\)

x^5+x+1=x^5-x^2+x^2+x+1

=x^2.(x^3-1)+(x^2+x+1)

=x^2.(x-1)(x^2+x+1)+(x^2+x+1)

=(x^2+x+1)(x^3-x^2+1)

\(A=\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{6}+\sqrt{8}+\sqrt{4}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

\(=\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{2}\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

\(=1+\sqrt{2}\)

\(A=\sqrt[3]{7+5\sqrt{2}}+\sqrt[3]{7-5\sqrt{2}}\)

\(\Rightarrow A^3=14+3\sqrt[3]{\left(7+5\sqrt{2}\right)\left(7-5\sqrt{2}\right)}\left(\sqrt[3]{7+5\sqrt{2}}+\sqrt[3]{7-5\sqrt{2}}\right)\)

<=>A³=14-3A

<=>A³+3A-14=0

<=>A³-4A+7A-14=0

<=>A.(A-2)(A+2)+7.(A-2)=0

<=>(A-2)(A²+9A-14)=0

<=>A=2(nhận)

Vậy A=2