\(3x=5y=10z\Leftrightarrow\dfrac{x}{\dfrac{1}{3}}=\dfrac{y}{\dfrac{1}{5}}=\dfrac{z}{\dfrac{1}{10}}\)

\(\Rightarrow\dfrac{x}{\dfrac{1}{3}}=\dfrac{2y}{\dfrac{2}{5}}=\dfrac{z}{\dfrac{1}{10}}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\dfrac{x}{\dfrac{1}{3}}=\dfrac{2y}{\dfrac{2}{5}}=\dfrac{z}{\dfrac{1}{10}}=\dfrac{x-2y-z}{\dfrac{1}{3}-\dfrac{2}{5}-\dfrac{1}{10}}=\dfrac{15}{-\dfrac{1}{6}}=-90\)

\(\Rightarrow\left\{{}\begin{matrix}x=-90.\dfrac{1}{3}=-30\\y=-90.\dfrac{1}{5}=-18\\z=-90.\dfrac{1}{10}=-9\end{matrix}\right.\)

\(\left(4x-5\right)\left(3x+2\right)=0\)

\(\)\(\Rightarrow\left[{}\begin{matrix}4x-5=0\\3x+2=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{4}\\x=-\dfrac{2}{3}\end{matrix}\right.\)

\(\dfrac{x+1}{2016}+\dfrac{x+2}{2015}=\dfrac{x+3}{2014}+\dfrac{x+4}{2013}\)

\(\Rightarrow\dfrac{x+1}{2016}+1+\dfrac{x+2}{2015}+1=\dfrac{x+3}{2014}+1+\dfrac{x+4}{2013}+1\)

\(\Rightarrow\dfrac{x+2017}{2016}+\dfrac{x+2017}{2015}=\dfrac{x+2017}{2014}+\dfrac{x+2017}{2013}\)

\(\Rightarrow\dfrac{x+2017}{2016}+\dfrac{x+2017}{2015}-\dfrac{x+2017}{2014}-\dfrac{x+2017}{2013}=0\)

\(\Rightarrow\left(x+2017\right)\left(\dfrac{1}{2016}+\dfrac{1}{2015}-\dfrac{1}{2014}-\dfrac{1}{2013}\right)=0\)

Vì \(\dfrac{1}{2016}+\dfrac{1}{2015}-\dfrac{1}{2014}-\dfrac{1}{2013}\ne0\)

Nên:

\(x+2017=0\Rightarrow x=-2017\)

Cx khá dễ:

\(x+y+13=2\left(2\sqrt{x}+3\sqrt{y}\right)\)

\(\Rightarrow x+y+13=4\sqrt{x}+6\sqrt{y}\)

\(\Rightarrow x+y+13-4\sqrt{x}-6\sqrt{y}=0\)

\(\Rightarrow x+y+4+9-4\sqrt{x}-6\sqrt{y}=0\)

\(\Rightarrow\left(x-4\sqrt{x}+4\right)+\left(y-6\sqrt{y}+9\right)=0\)

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

\(\left\{{}\begin{matrix}\left(\sqrt{x}-2\right)^2\ge0\\\left(\sqrt{y}-3\right)^2\ge0\end{matrix}\right.\) \(\Rightarrow\left(\sqrt{x}-2\right)^2+\left(\sqrt{y}-3\right)^2\ge0\)

Dấu "=" xảy ra khi:

\(\left\{{}\begin{matrix}\sqrt{x}=2\Rightarrow x=4\\\sqrt{y}=3\Rightarrow y=9\end{matrix}\right.\)

Đặt:

\(\dfrac{a}{b}=\dfrac{c}{d}=k\)

\(\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)

\(\left\{{}\begin{matrix}\dfrac{a+b}{c+d}=\dfrac{bk+b}{dk+d}=\dfrac{b\left(k+1\right)}{d\left(k+1\right)}=\dfrac{b}{d}\\\dfrac{a-2b}{c-2d}=\dfrac{bk-2b}{dk-2d}=\dfrac{b\left(k-2\right)}{d\left(k-2\right)}=\dfrac{b}{d}\end{matrix}\right.\)

Ta có điều phải chứng minh

\(A=1+3+3^2+3^3+...+3^{99}+3^{100}\)

\(3A=3+3^2+3^3+3^4+...+3^{100}+3^{101}\)

\(3A-A=\left(3+3^2+3^3+3^4+...+3^{100}+3^{101}\right)-\left(1+3+3^2+3^3+...+3^{99}\right)\)

\(2A=3^{101}-1\Leftrightarrow A=\dfrac{3^{101}-1}{2}\)

B đề sai

\(A=-\left(\dfrac{4}{9}x+\dfrac{2}{15}\right)^2+3\)

\(\left(\dfrac{4}{9}x+\dfrac{2}{15}\right)^2\ge0\Leftrightarrow-\left(\dfrac{4}{9}x+\dfrac{2}{15}\right)^2\le0\)

\(A=-\left(\dfrac{4}{9}x+\dfrac{2}{15}\right)^2+3\le3\)

Dấu "=" xảy ra khi:

\(-\left(\dfrac{4}{9}x+\dfrac{2}{15}\right)^2=0\Leftrightarrow x=-\dfrac{3}{10}\)

Mấy bài dễ tự làm nhé:D

1)

Đặt: \(\dfrac{a}{b}=\dfrac{c}{d}=k\Leftrightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)

\(\left\{{}\begin{matrix}\dfrac{a}{a+b}=\dfrac{bk}{bk+b}=\dfrac{bk}{b\left(k+1\right)}=\dfrac{k}{k+1}\\\dfrac{c}{c+d}=\dfrac{dk}{dk+d}=\dfrac{dk}{d\left(k+1\right)}=\dfrac{k}{k+1}\end{matrix}\right.\)

Ta có điều phải chứng minh

\(\left\{{}\begin{matrix}\dfrac{a}{a-b}=\dfrac{bk}{bk-b}=\dfrac{bk}{b\left(k-1\right)}=\dfrac{k}{k-1}\\\dfrac{c}{c-d}=\dfrac{dk}{dk-d}=\dfrac{dk}{d\left(k-1\right)}=\dfrac{k}{k-1}\end{matrix}\right.\)

Ta có điều phải chứng minh

*Thất vọng về you*

\(\dfrac{\left(-2\right)^n}{-128}=4\)

\(\Rightarrow\dfrac{\left(-2\right)^n}{\left(-2\right)^7}=\left(-2\right)^2\)

\(\Rightarrow\left(-2\right)^n=\left(-2\right)^7.\left(-2\right)^2\)

\(\Rightarrow\left(-2\right)^n=\left(-2\right)^9\)

2)\(\dfrac{\left(-\dfrac{1}{2}\right)^{2n}}{\left(-\dfrac{1}{2}\right)^n}=\dfrac{\left(-\dfrac{1}{2}\right)^n.\left(-\dfrac{1}{2}\right)^n}{\left(-\dfrac{1}{2}\right)^n}=\left(-\dfrac{1}{2}\right)^n\)

\(\left(x-5\right)\left(x+4\right)-\left(x-1\right)\left(x+3\right)=-2x^2\)

\(\Rightarrow\left[x\left(x+4\right)-5\left(x+4\right)\right]-\left[x\left(x+3\right)-1\left(x+3\right)\right]=-2x^2\)

\(\Rightarrow\left(x^2+4x-5x-20\right)-\left(x^2+3x-x-3\right)=-2x^2\)

\(\Rightarrow\left(x^2-x-20\right)-\left(x^2+2x-3\right)=-2x^2\)

\(\Rightarrow x^2-x-20-x^2-2x+3=-2x^2\)

\(\Rightarrow-x-17=-2x^2\)

Cạn....

\(x^3+x=0\)

\(\Rightarrow x\left(x^2+1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x^2+1=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x^2=-1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x\in\varnothing\end{matrix}\right.\)