![](https://rs.olm.vn/images/avt/0.png?1311)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
`\sqrt{[27(x-1)^2]/12} +3/2 - (x - 2)\sqrt{[50x^2]/[8(x-2)^2]}` `(1 < x < 2)`
`=\sqrt{[3(x-1)]^2 .3}/\sqrt{2^2 .3} + 3/2 - (x - 2) \sqrt{(5x)^2 . 2}/\sqrt{[2(x - 2)]^2 . 2}`
`=[3\sqrt{3}|x-1|]/[2\sqrt{3}]+3/2-(x-2)[5\sqrt{2}|x|]/[2\sqrt{2}|x-2|]`
`=[3(x-1)]/2+3/2-[5x(x-2)]/[2(2-x)]` (Vì `1 < x < 2`)
`=3/2x - 3/2 + 3/2 + 5/2x`
`=4x`
![](https://rs.olm.vn/images/avt/0.png?1311)
Vũ Minh Tuấn Lê Thị Thục Hiền @Nk>↑@ Băng Băng 2k6
\(\left(x+3\right)\sqrt{10-x^2}=x^2-x-12\) () (đk: \(-\sqrt{10}< x< \sqrt{10}\))
<=>\(\left(x+3\right)\sqrt{10-x^2}=x^2-4x+3x-12\)
<=> \(\left(x+3\right)\sqrt{10-x^2}=\left(x-4\right)\left(x+3\right)\)
<=> \(\left(x+3\right)\sqrt{10-x^2}-\left(x-4\right)\left(x+3\right)=0\)
<=> \(\left(x+3\right)\left(\sqrt{10-x^2}-x+4\right)=0\)
=>\(\left[{}\begin{matrix}x+3=0\\\sqrt{10-x^2}-x+4=0\end{matrix}\right.\) <=> \(\left[{}\begin{matrix}x=-3\left(tm\right)\\\sqrt{10-x^2}=x-4\end{matrix}\right.\) <=> \(\left[{}\begin{matrix}x=-3\\10-x^2=16-8x+x^2\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}x=-3\\0=6-8x+2x^2\end{matrix}\right.\) <=> \(\left[{}\begin{matrix}x=-3\\x^2-4x+3=0\end{matrix}\right.\) <=> \(\left[{}\begin{matrix}x=-3\\x^2-x-3x+3=0\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}x=-3\\\left(x-1\right)\left(x-3\right)=0\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}x=-3\left(tm\right)\\x=1\left(ktm\right)\\x=3\left(ktm\right)\end{matrix}\right.\)
Vậy pt (*) có nghiệm duy nhất x=-3
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(\sqrt{x+5}-\sqrt{x+2}\right)\left(1+\sqrt{x^2+7x+10}\right)=3\left(đk:x\ge-2\right)\)
Đặt \(a=\sqrt{x+5},b=\sqrt{x+2}\left(đk:a,b\ge0,a\ne b\right)\)
\(\Rightarrow\left\{{}\begin{matrix}ab=\sqrt{\left(x+5\right)\left(x+2\right)}=\sqrt{x^2+7x+10}\\a^2-b^2=x+5-x-2=3\end{matrix}\right.\)
PT trở thành: \(\left(a-b\right)\left(1+ab\right)=a^2-b^2\)
\(\Leftrightarrow\left(a-b\right)\left(ab+1\right)=\left(a-b\right)\left(a+b\right)\)
\(\Leftrightarrow\left(a-b\right)\left(ab+1-a-b\right)=0\)
\(\Leftrightarrow\left(a-b\right)\left(b-1\right)\left(a-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=b\left(loại\right)\\a=1\\b=1\end{matrix}\right.\)
+ Với a=1
\(\Rightarrow\sqrt{x+5}=1\Leftrightarrow x+5=1\Leftrightarrow x=-4\left(ktm\right)\)
+ Với b=1
\(\Rightarrow\sqrt{x+2}=1\Leftrightarrow x+2=1\Leftrightarrow x=-1\left(tm\right)\)
Vậy \(S=\left\{-1\right\}\)
Đặt \(\left\{{}\begin{matrix}\sqrt{x+5}=a\\\sqrt{x+2=b}\end{matrix}\right.\)
Thì được:
\(\left(a-b\right)\left(1+ab\right)=a^2-b^2\)
\(\Leftrightarrow\left(a-1\right)\left(b-1\right)\left(a-b\right)=0\)
Làm tiếp
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Giải phương trình : $\sqrt{x^{2}+5}+3x =\sqrt{x^{2}+12}+5$ - posted in Đại ... Giải. Dễ thấy, nếu x < 0: VT=√x2+5+3x<√x2+12<√x2+12+5 V T = x 2 + .... phương trình đã cho tương đương √x2+5+√x2+12=73x−5 x 2 + 5 + x 2 ...
x=-3 la nghiem
(chi tiet sau)
Đk:\(-\sqrt{10}\le x\le\sqrt{10}\)
\(\left(x+3\right)\sqrt{10-x^2}=x^2-x-12\)
\(\Leftrightarrow\left(x+3\right)\sqrt{10-x^2}=\left(x+3\right)\left(x-4\right)\)
\(\Leftrightarrow\left(x+3\right)\sqrt{10-x^2}-\left(x+3\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left[\sqrt{10-x^2}-\left(x-4\right)\right]=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+3=0\\\sqrt{10-x^2}-\left(x-4\right)=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=-3\\\sqrt{10-x^2}=x-4\left(\text{*}\right)\end{cases}}\)
Đk(*):\(x\ge4\). Bình phương 2 vế ta có:
\(10-x^2=x^2-8x+16\)
\(\Leftrightarrow x^2-4x+3=0\)
\(\Delta=\left(-4\right)^2-4\cdot1\cdot3=4\)
\(\Leftrightarrow x_{1,2}=\frac{4\pm\sqrt{4}}{2}\) \(\Rightarrow\orbr{\begin{cases}x_1=1\\x_2=3\end{cases}}\) (loại vì \(x\ge4\))
Vậy....