Sửa đề: \(B=\dfrac{x-3\sqrt{x}+4}{x-2}-\dfrac{1}{\sqrt{x}-2}\)

Ta có: \(B=\dfrac{x-3\sqrt{x}+4}{x-2}-\dfrac{1}{\sqrt{x}-2}\)

\(=\dfrac{x-3\sqrt{x}+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\dfrac{\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

\(=\dfrac{x-3\sqrt{x}+4-\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

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

\(x.\sqrt[3]{x}-22\sqrt[3]{x^2}+4=0\)

Đặt \(\sqrt[3]{x}\Rightarrow t\left(t\ge0\right)\)

Thì pt đã cho tương đương : 

\(t.x-t^2.22+4=0\)

Xét \(\Delta=x^2-4.\left(-22\right).4=x^2+352>0\)

nên pt có 2 nghiệm : \(t_1=\frac{-x+\sqrt{x^2+352}}{-44}=\sqrt[3]{x}\)easy :))

\(t_2=\frac{-x-\sqrt{x^2+352}}{-44}=\sqrt[3]{x}\)easy part 2 :0

Vậy nghiệm của pt trên là : ...

Chép sai đề kìa.

\(x\sqrt[3]{x}-22\sqrt[3]{x^2}+49=0\)

Đặt \(\sqrt[3]{x}=u\left(u\ge0\right)\)Ta có pt mới : \(xu-22u^2+49=0\)

Lịa có : \(\Delta=x^2-4.\left(-22\right).49=x+4322>0\)

Nên pt có 2 nghiệm phân biệt 

\(x_1=\frac{-x-\sqrt{4322}}{-44};x_2=\frac{-x+\sqrt{4322}}{-44}\)

\(\sqrt{3}cosx+2sin^2\left(\dfrac{x}{2}-\pi\right)=1\) 

\(\Leftrightarrow\sqrt{3}cosx+2sin^2\dfrac{x}{2}=1\)

\(\Leftrightarrow\sqrt{3}cosx-cosx=0\Leftrightarrow cosx=0\Leftrightarrow x=\dfrac{\pi}{2}+k\pi\) ( k thuộc Z )

Vậy ... 

22.

Nhận thấy \(cosx=0\) không phải nghiệm, chia 2 vế cho \(cos^2x\)

\(3tan^2x+2tanx-1=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{4}+k\pi\\x=arctan\left(\dfrac{1}{3}\right)+k\pi\end{matrix}\right.\)

Nghiệm dương nhỏ nhất của pt là: \(x=arctan\left(\dfrac{1}{3}\right)\)

22. PT đã cho tương đương

3 - 4cos²x + 2 sinxcosx = 0

⇔ 3 - 2 - 2cos2x + sin2x = 0

⇔ 1 - 2cos2x + sin2x = 0

⇔ 1 + sin2x = 2cos2x

⇔ sin\(\dfrac{\pi}{2}\) + sin2x = 2cos2x

⇔ \(2sin\left(\dfrac{\pi}{4}+x\right).cos\left(\dfrac{\pi}{4}-x\right)\) = 2cos2x

Do \(\left(\dfrac{\pi}{4}-x\right)+\left(\dfrac{\pi}{4}+x\right)=\dfrac{\pi}{2}\) 

⇒ \(sin\left(\dfrac{\pi}{4}+x\right)=cos\left(\dfrac{\pi}{4}-x\right)\)

Vậy sin²\(\left(x+\dfrac{\pi}{4}\right)\) = cos2x

Cái này là hiển nhiên ????

a) Ta có: \(\sqrt{x-2\sqrt{x-1}}-\sqrt{x-1}=1\)

\(\Leftrightarrow\left|\sqrt{x-1}-1\right|=\sqrt{x-1}+1\)

\(\Leftrightarrow\sqrt{x-1}=\sqrt{x-1}+1+1\)(Vô lý)

Vậy: \(S=\varnothing\)

b) Ta có: \(\sqrt{x^4+2x^2+1}=\sqrt{x^2+10x+25}-10x+22\)

\(\Leftrightarrow x^2+1=\left|x+5\right|-10x+22\)

\(\Leftrightarrow\left|x+5\right|=x^2+1+10x-22=x^2+10x-21\)

\(\Leftrightarrow\left[{}\begin{matrix}x+5=x^2+10x-21\left(x\ge-5\right)\\-x-5=x^2+10x-21\left(x< -5\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2+10x-21-x-5=0\\x^2+10x-21+x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x^2+9x-26=0\\x^2+11x-16=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-9+\sqrt{185}}{2}\\x=\dfrac{-11-\sqrt{185}}{2}\end{matrix}\right.\)

a , Ta có :

\(\Leftrightarrow\sqrt{7-x}=x-1\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-1\ge0\\7-x=x^2-2x+1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge1\\x^2-x-6=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge1\\\left[{}\begin{matrix}x=3\left(tm\right)\\x=-2\left(loại\right)\end{matrix}\right.\end{matrix}\right.\)

Vậy pt có nghiệm là x = 3

b , c , d , e , f tương tự

Đặt \(A=\sqrt[3]{22\sqrt{2}+25}-\sqrt[3]{22\sqrt{2}-25}\)

\(\Rightarrow A^3=50-3\sqrt[3]{\left(22\sqrt{2}+25\right)\left(22\sqrt{2}-25\right)}\left(\sqrt[3]{22\sqrt{2}+25}-\sqrt[3]{22\sqrt{2}-25}\right)\)

\(\Rightarrow A^3=50-3\sqrt[3]{\left(22\sqrt{2}+25\right)\left(22\sqrt{2}-25\right)}\cdot A\)

\(\Rightarrow A^3=50-3A\sqrt[3]{343}=50-21A\)

\(\Rightarrow A^3+21A-50=0\Leftrightarrow A^3-4A+25A-50=0\)

\(\Leftrightarrow\left(A-2\right)\left(A^2+2A+25\right)=0\)

\(\Leftrightarrow A=2\left(A^2+2A+25>0,\forall A\right)\)

\(\Rightarrow\sqrt[3]{22\sqrt{2}+25}-\sqrt[3]{22\sqrt{2}-25}=2\)

Tick nha bạn 😘

\(\Leftrightarrow sin\left(x+17^0+x-22^0\right)=\frac{\sqrt{2}}{2}\)

\(\Leftrightarrow sin\left(2x-5^0\right)=\frac{\sqrt{2}}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-5^0=45^0+k360^0\\2x-5^0=135^0+k360^0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=25^0+k180^0\\x=70^0+k180^0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=25^0\\x=70^0\end{matrix}\right.\)