giúp j vậy

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

\(\Leftrightarrow4-x^2=\left(2-x\right)^2\)

\(\Leftrightarrow4-x^2=4-8x+x^2\)

\(\Leftrightarrow4-x^2-4+8x-x^2=0\)

\(\Leftrightarrow8x-2x^2=0\)

\(\Leftrightarrow2x\left(4-x\right)=0\)

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

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

\(\Leftrightarrow1-x^2=\left(1-x\right)^2\)

\(\Leftrightarrow1-x^2=1-2x+x^2\)

\(\Leftrightarrow1-x^2-1+2x-x^2=0\)

\(\Leftrightarrow2x-2x^2=0\)

\(\Leftrightarrow2x\left(1-x\right)=0\)

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

_{Công viên nước ở Thái Bình thật đẹp.Có những cái cầu trượt cong queo trượt xuống hồ bơi.}

7. How often does he go to the library?
8. ....is the shortest student in his class.
9......is her address?
10... is shorter than Ba.

7 how often does he go to the library

8 nga is the shortest student in the class

9 what is her adress

10 hoa is shorter than ba

em cắt bớt đi

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) \(ĐK:x\ge-1\)

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

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

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

\(3,\sqrt{x^2-2x+1}=1\left(x\in R\right)\\ \Leftrightarrow\sqrt{\left(x-1\right)^2}=1\\ \Leftrightarrow\left|x-1\right|=1\Leftrightarrow\left[{}\begin{matrix}x-1=1\left(x\ge1\right)\\x-1=-1\left(x< 1\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\left(tm\right)\\x=0\left(tm\right)\end{matrix}\right.\)

\(4,ĐK:x\in R\\ PT\Leftrightarrow\sqrt{\left(x-2\right)^2}=1\\ \Leftrightarrow\left|x-2\right|=1\Leftrightarrow\left[{}\begin{matrix}x-2=1\left(x\ge2\right)\\x-2=-1\left(x< 2\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\left(tm\right)\\x=1\left(tm\right)\end{matrix}\right.\)

\(5,ĐK:x\in R\\ PT\Leftrightarrow\left|2x-1\right|=\left|x+4\right|\\ \Leftrightarrow\left[{}\begin{matrix}2x-1=x+4\\1-2x=x+4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-1\end{matrix}\right.\)

\(6,ĐK:x\ge-2\\ PT\Leftrightarrow5\sqrt{x+2}-3\sqrt{x+2}-\sqrt{x+2}=\sqrt{x+7}\\ \Leftrightarrow\sqrt{x+2}=\sqrt{x+7}\Leftrightarrow x+2=x+7\Leftrightarrow0x=5\Leftrightarrow x\in\varnothing\)

\(7,ĐK:x\ge-1\\ PT\Leftrightarrow5\sqrt{x+2}+3\sqrt{x+1}=4\sqrt{x+1}+4\sqrt{x+2}\\ \Leftrightarrow\sqrt{x+2}=\sqrt{x+1}\\ \Leftrightarrow x+2=x+1\\ \Leftrightarrow0x=-1\Leftrightarrow x\in\varnothing\)