Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Hung nguyen, Trần Thanh Phương, Sky SơnTùng, @tth_new, @Nguyễn Việt Lâm, @Akai Haruma, @No choice teen
help me, pleaseee
Cần gấp lắm ạ!
a) \(\sqrt{x^2-9}-\sqrt{4x-12}=0\) ĐK: \(x\ge3\)
\(\Leftrightarrow\sqrt{\left(x-3\right)\left(x+3\right)}-2\sqrt{x-3}=0\)
\(\Leftrightarrow\sqrt{x-3}\left(\sqrt{x+3}-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-3}=0\\\sqrt{x+3}-2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)
Vậy x = 3
b) \(\sqrt{1-x}+\sqrt{x}=1\) ĐK: \(0\le x\le1\)
\(\Leftrightarrow1-x+x+2\sqrt{x\left(1-x\right)}=1\)
\(\Leftrightarrow x\left(1-x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\) (Nhận)
c) \(\sqrt{x+3}+\sqrt{x+8}=5\) ĐK: \(x\ge-3\)
Đặt \(\left\{{}\begin{matrix}a=\sqrt{x+3}\ge0\\b=\sqrt{x+8}\ge0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}a+b=5\\b^2-a^2=5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a+b=5\\b-a=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=2\\b=3\end{matrix}\right.\)
\(\Leftrightarrow x=1\) (Nhận)
d) \(\sqrt{16-32x}-\sqrt{12x}=\sqrt{3x}+\sqrt{9-18x}\) ĐK: \(-\dfrac{1}{2}\le x\le0\)
\(\Leftrightarrow4\sqrt{1-2x}-2\sqrt{3x}=\sqrt{3x}+3\sqrt{1-2x}\)
\(\Leftrightarrow\sqrt{1-2x}=3\sqrt{3x}\)
\(\Leftrightarrow1-2x=27x\)
\(\Leftrightarrow x=\dfrac{1}{29}\) (Nhận)
Làm hơi tắt xíu, có gì ko hiểu cmt nha :>
\(a.\sqrt{x-1}=3\left(ĐK:x\ge1\right)\Leftrightarrow x-1=9\Leftrightarrow x=10\)
\(b.\sqrt{x^2-4x+4}=2\\ \Leftrightarrow\sqrt{\left(x-2\right)^2}=2\\ \Leftrightarrow\left|x-2\right|=2\\ \Leftrightarrow\left[{}\begin{matrix}x-2=2\left(x\ge2\right)\\2-x=2\left(x< 2\right)\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=4\\x=0\end{matrix}\right.\)
\(c.\sqrt{25x^2-10x+1}=4x-9\\ \Leftrightarrow\sqrt{\left(5x-1\right)^2}=4x-9\\ \Leftrightarrow\left|5x-1\right|=4x-9\\\Leftrightarrow \left[{}\begin{matrix}5x-1=4x-9\left(x\ge\frac{1}{5}\right)\\1-5x=4x-9\left(x< \frac{1}{5}\right)\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-8\left(ktm\right)\\x=\frac{10}{9}\left(ktm\right)\end{matrix}\right.\)
\(d.\sqrt{x^2+2x+1}=\sqrt{x+1}\left(ĐK:x\ge-1\right)\\ \Leftrightarrow x^2+2x+1=x+1\\ \Leftrightarrow x^2+x=0\Leftrightarrow x\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)
e. ĐK: \(\left[{}\begin{matrix}x\ge3\\x\le-3\end{matrix}\right.\)
\(\sqrt{x^2-9}+\sqrt{x^2-6x+9}=0\\ \Leftrightarrow\sqrt{\left(x-3\right)\left(x+3\right)}+\sqrt{\left(x-3\right)^2}=0\\ \Leftrightarrow\sqrt{x-3}\left(\sqrt{x+3}+\sqrt{x-3}\right)=0\\ \Leftrightarrow\sqrt{x-3}=0\\ \Leftrightarrow x-3=0\Leftrightarrow x=3\)
Câu cuối chưa nghĩ ra, sorry :<
Liên hợp:v
a) ĐK: \(x\ge-2\)
PT<=> \(\sqrt{x+5}-2+\sqrt{x+2}-1+2\left(x+1\right)=0\)
\(\Leftrightarrow\frac{x+1}{\sqrt{x+5}+2}+\frac{x+1}{\sqrt{x+2}+1}+2\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(\frac{1}{\sqrt{x+5}+2}+\frac{1}{\sqrt{x+2}+1}+2\right)=0\)
Cái ngoặc to nhìn sơ qua cũng thấy nó >0 :v
Do đó x = -1
Vậy...
P/s: cô @Akai Haruma check giúp em ạ!
Nguyễn Việt Lâm, svtkvtm, Trần Thanh Phương, Phạm Hoàng Hải Anh, DƯƠNG PHAN KHÁNH DƯƠNG, @Akai Haruma
Bài 1: Giải phương trình
a) ĐKXĐ: \(x\ge3\)
Ta có: \(\sqrt{100\cdot\left(x-3\right)}=\sqrt{20}\)
\(\Leftrightarrow\left|100\cdot\left(x-3\right)\right|=\left|20\right|\)
\(\Leftrightarrow100\cdot\left|x-3\right|=20\)
\(\Leftrightarrow\left|x-3\right|=\frac{1}{5}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=\frac{1}{5}\\x-3=-\frac{1}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{16}{5}\left(nhận\right)\\x=\frac{14}{5}\left(loại\right)\end{matrix}\right.\)
Vậy: \(S=\left\{\frac{16}{5}\right\}\)
b) Ta có: \(\sqrt{\left(x-3\right)^2}=7\)
\(\Leftrightarrow\left|x-3\right|=7\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=7\\x-3=-7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=10\\x=-4\end{matrix}\right.\)
Vậy: S={10;-4}
c) Ta có: \(\sqrt{4x^2+4x+1}=6\)
\(\Leftrightarrow\sqrt{\left(2x+1\right)^2}=6\)
\(\Leftrightarrow\left|2x+1\right|=6\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+1=6\\2x+1=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=5\\2x=-7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{5}{2}\\x=\frac{-7}{2}\end{matrix}\right.\)
Vậy: \(S=\left\{\frac{5}{2};\frac{-7}{2}\right\}\)
Bài 1:
b: \(\Leftrightarrow2+\sqrt{3x-5}=x+1\)
\(\Leftrightarrow\sqrt{3x-5}=x-1\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2-2x+1=3x-5\\x>=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x^2-5x+6=0\\x>=1\end{matrix}\right.\Leftrightarrow x\in\left\{2;3\right\}\)
c: \(\Leftrightarrow5x+7=16\left(x+3\right)\)
=>16x+48=5x+7
=>11x=-41
hay x=-41/11
a, ĐKXĐ : \(x-1\ge0\)
=> \(x\ge1\)
Ta có : \(x+2\sqrt{x-1}-9=0\)
=> \(x-1+2\sqrt{x-1}+1-9=0\)
=> \(\left(\sqrt{x-1}+1\right)^2=9\)
=> \(\left[{}\begin{matrix}\sqrt{x-1}+1=3\\\sqrt{x-1}+1=-3\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}\sqrt{x-1}=2\\\sqrt{x-1}=-4\left(VL\right)\end{matrix}\right.\)
=> \(x-1=4\)
=> \(x=5\left(TM\right)\)
Vậy ...
b, ĐKXĐ : \(x-2\ge0\)
=> \(x\ge2\)
Ta có : \(3x-\sqrt{x-2}-8=0\)
=> \(x-\frac{2.1}{6}\sqrt{x-2}-\frac{8}{3}=0\)
=> \(x-2-2.\frac{1}{6}\sqrt{x-2}+\frac{1}{36}-\frac{25}{36}=0\)
=> \(\left(\sqrt{x-2}-\frac{1}{6}\right)^2=\frac{25}{36}\)
=> \(\left[{}\begin{matrix}\sqrt{x-2}-\frac{1}{6}=\frac{5}{6}\\\sqrt{x-2}-\frac{1}{6}=-\frac{5}{6}\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}\sqrt{x-2}=1\\\sqrt{x-2}=-\frac{2}{3}\left(L\right)\end{matrix}\right.\)
=> \(x-2=1\)
=> \(x=3\left(TM\right)\)
Vậy ....