ta có \(\sin\alpha:\cos\alpha=\frac{đ}{h}:\frac{k}{h}=\frac{đ}{h}.\frac{h}{k}=\frac{đ}{k}=\tan\alpha \left(đpcm\right)\)

Góc B bằng alpha suy ra:

\(\frac{sin\alpha}{cos\alpha}=\frac{\frac{AC}{BC}}{\frac{AB}{BC}}=\frac{AC}{AB}=tan\alpha\)(đpcm)

a) 12 . ( x - 1 ) = 0

            x - 1   = 0 : 12 

            x - 1   = 0

              x      = 0 + 1

              x      = 1

b) ( 6x - 39 ) : 3 = 201

            6x - 39 = 201 . 3 

            6x - 39 = 603

               6x    = 603 + 39

               6x    = 642

                x     = 642 : 6 

                x     = 107

c) 23 + 3x = 5⁶ : 5³

    23 + 3x = 5³ 

    23 + 3x = 125

            3x = 125 - 23

            3x = 102

            x  = 102 : 3 

            x  = 34

Các phần d , e , f Đỗ Ngọc Hoàng Hải làm tương tự như phần trên .

a/x =1 nha

b/(6x-39):3=201

   6x-39    =201x3

   6x-39    =603

   6x         = 603+39

   6x = 642

   6= 642:6=107

c/ 23+3.x=125

         3x= 125-23=102

         x= 102:3=34

d/ 541+(281-x)=735

            281-x= 735-541=194

             x=281-194=87

e/ 9x+2=20

    9x=20-2

    9x=18

    x=18:9=9

f/71+(26-3x):5=75

       (26-3x):5=75-71=4

       26-3x=4x5=20

           3x=26-20=6

            x=6:3=2

tk nha ^^

=\(\frac{\sqrt{4+2\sqrt{3}}}{2\sqrt{2}}:\left(\frac{\sqrt{4+2\sqrt{3}}}{2\sqrt{2}}-\frac{2}{\sqrt{6}}+\frac{\sqrt{4+2\sqrt{3}}}{2\sqrt{6}}\right)\)

=\(\frac{\sqrt{\left(\sqrt{3}+1\right)^2}}{2\sqrt{2}}:\left(\frac{\sqrt{\left(\sqrt{3}+1\right)^2}}{2\sqrt{2}}-\frac{2}{\sqrt{6}}+\frac{\sqrt{\left(\sqrt{3}+1\right)^2}}{2\sqrt{6}}\right)\)

=\(\frac{\sqrt{3}+1}{2\sqrt{2}}:\left(\frac{\sqrt{3}+1}{2\sqrt{2}}-\frac{2}{\sqrt{6}}+\frac{\sqrt{3}+1}{2\sqrt{6}}\right)\)

=\(\frac{\sqrt{3}+1}{2\sqrt{2}}:\frac{\sqrt{3}.\left(\sqrt{3}+1\right)-2.2+\sqrt{3}+1}{2\sqrt{6}}\)

=\(\frac{\sqrt{3}+1}{2\sqrt{2}.}.\frac{2\sqrt{6}}{2\sqrt{3}}=\frac{\sqrt{3}+1}{2}\)

\(A=x^2-5\ge-5,\forall x\)Vậy minA = -5 khi x = 0

\(B=\sqrt{x^2+4}\ge\sqrt{4}=2,\forall x\)Vậy minB = 2 khi x = 0

\(C=\sqrt{2x^2+3}\ge\sqrt{3},\forall x\)Vậy minC = \(\sqrt{3}\)khi x = 0

a)Đk:\(x\ge\frac{1}{2}\)

\(pt\Leftrightarrow4x^2-12x+4+4\sqrt{2x-1}=0\)

\(\Leftrightarrow\left(2x-1\right)^2-4\left(2x-1\right)-1+4\sqrt{2x-1}=0\)

Đặt \(t=\sqrt{2x-1}>0\Rightarrow\hept{\begin{cases}t^2=2x-1\\t^4=\left(2x-1\right)^2\end{cases}}\)

\(t^4-4t^2+4t-1=0\)

\(\Leftrightarrow\left(t-1\right)^2\left(t^2+2t-1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}t-1=0\\t^2+2t-1=0\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}t=1\\t=\sqrt{2}-1\end{cases}\left(t>0\right)}\)

\(\Rightarrow\orbr{\begin{cases}x=1\\x=2-\sqrt{2}\end{cases}}\) là nghiệm thỏa pt

ĐK \(x\ge-2\)

\(\sqrt{x+2}=2-x^2\)\(\Rightarrow x+2=\left(2-x^2\right)^2\)với ĐK \(2-x^2\ge0\Rightarrow-\sqrt{2}< x< \sqrt{2}\)

\(\Rightarrow x+2=4-4x^2-x^4\)\(\Rightarrow-x^4-4x^2-x+2=0\)

\(\Rightarrow x^2.\left(x^2-4\right)-\left(x-2\right)=0\Rightarrow\left(x-2\right)\left(x^2\left(x+2\right)-1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-2=0\\x^2\left(x+2\right)-1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=2\\x^3+2x^2-1=0\left(1\right)\end{cases}}}\)

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

\(\Rightarrow\orbr{\begin{cases}x+1=0\\x^2+x-1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-1\\x^2+x-1=0\left(2\right)\end{cases}}}\)

\(\left(2\right)\Leftrightarrow\) \(x^2+x-1=0\Rightarrow\orbr{\begin{cases}x=\frac{-1-\sqrt{5}}{2}\\x=\frac{-1+\sqrt{5}}{2}\end{cases}}\)

Kết hợp ĐK \(\hept{\begin{cases}x\ge-2\\-\sqrt{2}< x< \sqrt{2}\end{cases}}\) ta thấy \(x=-1\) hoặc \(x=\frac{-1+\sqrt{5}}{2}\)thỏa mãn 

Vậy \(x=-1\)hoặc \(x=\frac{-1+\sqrt{5}}{2}\)

a) sửa đề:  \(a^2+b^2\ge\frac{1}{2}\left(a+b\right)^2\)

\(\Leftrightarrow2\left(a^2+b^2\right)\ge\left(a+b\right)^2\)

\(\Leftrightarrow2a^2+2b^2-a^2-2ab-b^2\ge0\)

\(\Leftrightarrow a^2-2ab+b^2\ge0\)

\(\Leftrightarrow\left(a-b\right)^2\ge0\)  (luôn đúng)

Đẳng thức xảy ra  \(\Leftrightarrow\)  a = b

b) Đề hỏi gì vậy bn?

sịt cái đề sai hết