cái thứ 2 em tải hình xuống đề phòng hình 1 mất ạ

1) \(\sqrt{2x-5}=7\)

\(\left(\sqrt{2x-5}\right)^2=7^2\)

\(2x-5=49\)

\(2x=54\)

\(x=27\)

2) \(3+\sqrt{x-2}=4\)

\(\sqrt{x-2}=1\)

\(\left(\sqrt{x-2}\right)^2=1^2\)

\(x-2=1\)

\(x=3\)

1) \(\sqrt{2x-5}=7\left(đk:x\ge\dfrac{5}{2}\right)\)

\(\Leftrightarrow2x-5=49\Leftrightarrow2x=54\Leftrightarrow x=27\left(tm\right)\)

2) \(3+\sqrt{x-2}=4\left(đk:x\ge2\right)\)

\(\Leftrightarrow\sqrt{x-2}=1\Leftrightarrow x-2=1\Leftrightarrow x=3\)

3) \(\Leftrightarrow\sqrt{\left(x-1\right)^2}=1\Leftrightarrow\left|x-1\right|=1\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x-2=1\\x-2=-1\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)

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

\(\Leftrightarrow\left|2x-1\right|=\left|x+4\right|\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-1=x+4\\2x-1=-x-4\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-1\end{matrix}\right.\)

6) \(ĐK:x\ge-2\)

 \(\Leftrightarrow5\sqrt{x+2}-3\sqrt{x+2}-\sqrt{x+2}=\sqrt{x+7}\)

\(\Leftrightarrow\sqrt{x+2}=\sqrt{x+7}\)

\(\Leftrightarrow x+2=x+7\Leftrightarrow2=7\left(VLý\right)\)

Vậy \(S=\varnothing\)

7) \(ĐK:x\ge-1\)

\(\Leftrightarrow5\sqrt{2x+1}+3\sqrt{x+1}=4\sqrt{x+1}+4\sqrt{2x+1}\)

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

\(\Leftrightarrow2x+1=x+1\Leftrightarrow x=0\left(tm\right)\)

556667576

Bài 1: 

\(\sin\widehat{A}=\dfrac{BC}{BA}\)

\(\cos\widehat{A}=\dfrac{CA}{AB}\)

\(\tan\widehat{A}=\dfrac{BC}{CA}\)

\(\cot\widehat{A}=\dfrac{CA}{BC}\)

cô hoặc cj j ơi em đang cần 8 bài ạ 

\(x+y=2\Rightarrow y=2-x\)

\(xy=x.\left(2-x\right)=2x-x^2=-\left(x^2-2x\right)\)

                                                    \(=-\left(x^2-2x+1-1\right)=-\left(x-1\right)^2+1=1-\left(x-1\right)^2\le1\)

=> đpcm

( Dấu "=" xảy ra <=> x = 1 => y = 2 - x = 2 - 1 = 1 )

Bạn có cách giải bằng hình ko ạ

Bài 3:

a: \(M=\frac{x+12}{x-4}+\frac{1}{\sqrt{x}+2}-\frac{4}{\sqrt{x}-2}\)

\(=\frac{x+12+\sqrt{x}-2-4\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)

\(=\frac{x+\sqrt{x}+10-4\sqrt{x}-8}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}=\frac{x-3\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)

\(=\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}=\frac{\sqrt{x}-1}{\sqrt{x}+2}\)

b: Đặt \(P=\frac{1}{M}\)

\(=1:\frac{\sqrt{x}-1}{\sqrt{x}+2}=\frac{\sqrt{x}+2}{\sqrt{x}-1}\)

Để P là số nguyên thì \(\sqrt{x}+2\) ⋮\(\sqrt{x}-1\)

=>\(\sqrt{x}-1+3\) ⋮\(\sqrt{x}-1\)

=>3⋮\(\sqrt{x}-1\)

=>\(\sqrt{x}-1\in\left\lbrace1;-1;3;-3\right\rbrace\)

=>\(\sqrt{x}\in\left\lbrace2;0;4;-2\right\rbrace\)

=>\(\sqrt{x}\in\left\lbrace0;2;4\right\rbrace\)

=>x∈{0;4;16}

Kết hợp ĐKXĐ, ta được: x∈{0;16}

c: \(M-1=\frac{\sqrt{x}-1}{\sqrt{x}+2}-1=\frac{\sqrt{x}-1-\sqrt{x}-2}{\sqrt{x}+2}=\frac{-3}{\sqrt{x}+2}<0\)

=>M<1

d: \(M^2=-M\)

=>M(M+1)=0

=>M=0 hoặc M=-1

=>\(\left[\begin{array}{l}\frac{\sqrt{x}-1}{\sqrt{x}+2}=0\\ \frac{\sqrt{x}-1}{\sqrt{x}+2}=-1\end{array}\right.\Rightarrow\left[\begin{array}{l}\sqrt{x}-1=0\\ \sqrt{x}-1=-\sqrt{x}-2\end{array}\right.\)

=>\(\left[\begin{array}{l}\sqrt{x}=1\\ 2\sqrt{x}=-1\end{array}\right.\Rightarrow\sqrt{x}=1\)

=>x=1(nhận)

Bài 4:

a: \(\frac{x+2}{x\sqrt{x}-1}+\frac{\sqrt{x}}{x+\sqrt{x}+1}+\frac{1}{1-\sqrt{x}}\)

\(=\frac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\frac{\sqrt{x}}{x+\sqrt{x}+1}-\frac{1}{\sqrt{x}-1}\)

\(=\frac{x+2+\sqrt{x}\left(\sqrt{x}-1\right)-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)

\(=\frac{-\sqrt{x}+1+\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}=\frac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}=\frac{\sqrt{x}-1}{x+\sqrt{x}+1}\)

Ta có: \(P=\left(\frac{x+2}{x\sqrt{x}-1}+\frac{\sqrt{x}}{x+\sqrt{x}+1}+\frac{1}{1-\sqrt{x}}\right):\frac{\sqrt{x}-1}{2}\)

\(=\frac{\sqrt{x}-1}{x+\sqrt{x}+1}\cdot\frac{2}{\sqrt{x}-1}=\frac{2}{x+\sqrt{x}+1}\)

b: \(x+\sqrt{x}+1=\sqrt{x}\left(\sqrt{x}+1\right)+1>0\forall x\) thỏa mãn ĐKXĐ

2>0

Do đó: \(P=\frac{2}{x+\sqrt{x}+1}>0\forall x\) thỏa mãn ĐKXĐ

Bài 1:

1: Thay x=9 vào A, ta được:

\(A=\frac{\sqrt9+1}{\sqrt9-1}=\frac{3+1}{3-1}=\frac42=2\)

2:

a: \(\frac{x-2}{x+2\sqrt{x}}+\frac{1}{\sqrt{x}+2}\)

\(=\frac{x-2}{\sqrt{x}\left(\sqrt{x}+2\right)}+\frac{1}{\sqrt{x}+2}=\frac{x+\sqrt{x}-2}{\sqrt{x}\left(\sqrt{x}+2\right)}\)

\(=\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}+2\right)}=\frac{\sqrt{x}-1}{\sqrt{x}}\)

Ta có: \(P=\left(\frac{x-2}{x+2\sqrt{x}}+\frac{1}{\sqrt{x}+2}\right)\cdot\frac{\sqrt{x}+1}{\sqrt{x}-1}\)

\(=\frac{\sqrt{x}-1}{\sqrt{x}}\cdot\frac{\sqrt{x}+1}{\sqrt{x}-1}=\frac{\sqrt{x}+1}{\sqrt{x}}\)

b: \(2P=2\sqrt{x}+5\)

=>\(\frac{2\sqrt{x}+2}{\sqrt{x}}=2\sqrt{x}+5\)

=>\(2x+5\sqrt{x}=2\sqrt{x}+2\)

=>\(2x+3\sqrt{x}-2=0\)

=>\(2x+4\sqrt{x}-\sqrt{x}-2=0\)

=>\(\left(\sqrt{x}+2\right)\left(2\sqrt{x}-1\right)=0\)

=>\(2\sqrt{x}-1=0\)

=>\(2\sqrt{x}=1\)

=>\(\sqrt{x}=\frac12\)

=>x=1/4(nhận)

dạng này dễ mà bạn 

bạn tìm ĐK, đối chiếu giá trị với ĐK thấy thỏa mãn rồi thay vô 

toàn SCP nên tính cũng đơn giản:)

1) Thay x = 64 (TMĐK ) vào A, có :

           A = \(\frac{\sqrt{64}}{\sqrt{64}-2}\)=\(\frac{4}{3}\)

     Vậy A = \(\frac{4}{3}\)khi x = 64

2)  Thay x = 36 ( TMĐK ) vào A, có

        A =\(\frac{\sqrt{36}+4}{\sqrt{36}+2}\)=\(\frac{5}{4}\)

     Vậy A =\(\frac{5}{4}\)khi x = 36

3)   Thay x=9 (TMĐK  ) vào A, có :

         A= \(\frac{\sqrt{9}-5}{\sqrt{9}+5}\)=  \(\frac{-1}{4}\)

     Vậy A=\(\frac{-1}{4}\)khi x = 9

4)   Thay x = 25( TMĐK ) vào A có:

         A =\(\frac{2+\sqrt{25}}{\sqrt{25}}\)=\(\frac{7}{5}\)

      Vậy A=\(\frac{7}{5}\) khi x = 25

P₁ = (\(\frac{1}{\sqrt{x}}+\frac{\sqrt{x}}{\sqrt{x}+1}\)) : \(\frac{\sqrt{x}}{x+\sqrt{x}}\)= \(\frac{\sqrt{x}+1+x}{\sqrt{x}\left(\sqrt{x}+1\right)}\):\(\frac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\)=\(\frac{x+\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}\).

(\(\sqrt{x}+1\)) =\(\frac{x+\sqrt{x}+1}{\sqrt{x}}\)(ĐKXĐ : x > 0 )

P₂ =\(\frac{\sqrt{x}}{\sqrt{x}-1}+\frac{3}{\sqrt{x}+1}-\frac{6\sqrt{x}-4}{x-1}\)=\(\frac{\sqrt{x}\left(\sqrt{x}+1\right)+3\left(\sqrt{x}-1\right)-6\sqrt{x}+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)= \(\frac{x+\sqrt{x}+3\sqrt{x}-3-6\sqrt{x}+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)= \(\frac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)=\(\frac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)=\(\frac{\sqrt{x}-1}{\sqrt{x}+1}\)

(ĐKXĐ: x\(\ge\)0,  x\(\ne\)1)

Độ dài cạnh đối diện với góc 30 độ là 9cm

Dạ e cảm ơn