tớ chưa học

Xét cấp số cộng 1, 6, 11, ..., 96.

Ta có: 96 = 1 + 5(n − 1) ⇒ n = 20

Suy ra

Và 2x.20 + 970 = 1010

Từ đó x = 1

a: Ta sẽ có hình vẽ sau:

Đặt \(x=\widehat{B}\)

sin x=sin B=AC/BC

cosx=cosB=AB/BC

\(tanx=tanB=\dfrac{AC}{AB}=\dfrac{sinx}{cosx}\)

=>\(tan^2x=\dfrac{sin^2x}{cos^2x}\)

b: \(cot^2x=\dfrac{1}{tan^2x}=1:\dfrac{sin^2x}{cos^2x}=\dfrac{cos^2x}{sin^2x}\)

Ghi lại đề cho đúng đi bạn.

Sửa đề :

Tìm GTNN của \(A=\frac{x^2-2x+2011}{x^2}\)

Giải :

\(A=\frac{x^2-2x+2011}{x^2}\)

\(A=\frac{x^2}{x^2}-\frac{2x}{x^2}+\frac{2011}{x^2}\)

\(A=1-\frac{2}{x}+\frac{2011}{x^2}\)

\(A=1-2\cdot\frac{1}{x}+2011\cdot\left(\frac{1}{x}\right)^2\)

Đặt \(\frac{1}{x}=a\)

\(A=1-2a+2011a^2\)

\(A=2011\left(a^2-\frac{2}{2011}a+\frac{1}{2011}\right)\)

\(A=2011\left(a^2-2\cdot a\cdot\frac{1}{2011}+\frac{1}{2011^2}+\frac{2010}{4044121}\right)\)

\(A=2011\left[\left(a-\frac{1}{2011}\right)^2+\frac{2010}{4044121}\right]\)

\(A=2011\left(a-\frac{1}{2011}\right)^2+\frac{2010}{2011}\ge\frac{2010}{2011}\forall a\)

Dấu "=" xảy ra \(\Leftrightarrow a=\frac{1}{2011}\)

Thay a ta có : \(\frac{1}{x}=\frac{1}{2011}\)

\(\Rightarrow x=2011\)

Vậy \(A_{min}=\frac{2010}{2011}\Leftrightarrow x=2011\)

a) x² - x = 0 <=> x(x - 1) = 0 <=> x = 0 hoặc x - 1 = 0 <=> x = 0 hoặc x = 1

Vậy : S = {0; 1}.

b) x² - 2x = 0 <=> x(x - 2) <=> x = 0 hoặc x - 2 = 0 <=> x = 0 hoặc x = 2

Vậy : S = {0; 2).

(Bài này dễ mà)

Ta có : \(x^2+y^2-2x+4y+1\)

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

\(A=\left(x-1\right)^2+\left(y+2\right)^2-4\)

Vì \(\left(x-1\right)^2+\left(y+2\right)^2\ge0\forall x,y\in R\)

Nên : \(A=\left(x-1\right)^2+\left(y+2\right)^2-4\ge-4\forall x,y\in R\)

Vậy \(A_{min}=-4\) khi x = 1 và y = -2

\(\left(2x-1\right)^2-\left(x+3\right)^2=0\)

\(=>\left(2x-1+x+3\right)\left(2x-1-x-3\right)=0\)

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

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

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

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

\(=>x\in\left\{\dfrac{-2}{3};4\right\}\)

\(\left(2x-1\right)^2-\left(x+3\right)^2=0\)(sửa đề)

\(\Leftrightarrow\left(2x-1-x-3\right)\left(2x-1+x+3\right)=0\)

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

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