các bạn làm phép tính ra đi

Phương trình  sin 2 x + cos x = 0 có tổng các nghiệm trong khoảng 0 ; 2 π  bằng

A. 2 π

B.  3 π

C.  5 π

D.  6 π

tam giác ABC vuông tại A, đường cao AH

\(\Rightarrow BC.BH=AB^2\)

\(\Leftrightarrow BC\left(BC-CH\right)=AB^2\)

\(\Leftrightarrow BC^2-3,2BC-9=0\)

\(\Leftrightarrow BC^2+1,8BC-5BC-9=0\)

\(\Leftrightarrow BC\left(BC+1,8\right)-5\left(BC+1,8\right)=9\)

\(\Leftrightarrow\left(BC+1,8\right)\left(BC-5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}BC=-1,8\left(loại\right)\\BC=5\left(TM\right)\end{matrix}\right.\)

=> AC=\(\sqrt{BC^2-AB^2}=4\left(cm\right)\)

Có : AH.BC=AB.AC

=> AH=AB.AC/BC=3.4/5=2,4(cm)

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

\(\Leftrightarrow x^2+x=y^2+1\)

\(\Leftrightarrow4x^2+4x=4y^2+4\)

\(\Leftrightarrow4x^2+4x+1=4y^2+5\)

\(\Leftrightarrow\left(2x+1\right)^2-\left(2y\right)^2=5\)

\(\left(2x+1-2y\right)\left(2x+1+2y\right)=5=1.5=5.1=\left(-1\right).\left(-5\right)=\left(-5\right).\left(-1\right)\)

\(\left[{}\begin{matrix}\left\{{}\begin{matrix}2x+1-2y=1\\2x+1+2y=5\end{matrix}\right.\\\left\{{}\begin{matrix}2x+1-2y=5\\2x+1+2y=1\end{matrix}\right.\\\left\{{}\begin{matrix}2x+1-2y=-1\\2x+1+2y=-5\end{matrix}\right.\\\left\{{}\begin{matrix}2x+1-2y=-5\\2x+1+2y=-1\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=1\\y=1\end{matrix}\right.\\\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\\\left\{{}\begin{matrix}x=-2\\y=-1\end{matrix}\right.\\\left\{{}\begin{matrix}x=-2\\y=1\end{matrix}\right.\end{matrix}\right.\)

Vậy \(\left\{\left(x;y\right)\right\}=\left\{\left(1;1\right);\left(1;-1\right);\left(-2;-1\right);\left(-2;1\right)\right\}\)

20

Tik cho mk nha.........cảm ơn nhiều

có phải ở trong violympic ko?

\(\dfrac{a-b}{\sqrt{a}-\sqrt{b}}-\dfrac{\sqrt{a^3}-\sqrt{b^3}}{\sqrt{a}-\sqrt{b}}\)

\(=\dfrac{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{a}-\sqrt{b}}-\dfrac{\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)}{\sqrt{a}-\sqrt{b}}\)

\(=\sqrt{a}+\sqrt{b}-\left(a+\sqrt{ab}+b\right)\)

\(=\sqrt{a}+\sqrt{b}-a-\sqrt{ab}-b\)

Đề có sai không vậy bạn?

Phải là \(4\left(\sqrt{5}+1\right)\) chứ

Ta có:

\(x=\sqrt[3]{5\left(\sqrt{6}+1\right)}-\sqrt[3]{5\left(\sqrt{6}-1\right)}\)

\(\Leftrightarrow x^3=5\left(\sqrt{6}+1\right)-5\left(\sqrt{6}-1\right)-3.\sqrt[3]{5\left(\sqrt{6}+1\right)5\left(\sqrt{6}-1\right)}.x\)

\(\Leftrightarrow x^3=10-3.\sqrt[3]{5^2\left(6-1\right)}x\)

\(\Leftrightarrow x^3=10-15x\)

\(\Leftrightarrow x^3+15x=10\)