Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Câu 4:
Giả sử điều cần chứng minh là đúng
\(\Rightarrow x=y\), thay vào điều kiện ở đề bài, ta được:
\(\sqrt{x+2014}+\sqrt{2015-x}-\sqrt{2014-x}=\sqrt{x+2014}+\sqrt{2015-x}-\sqrt{2014-x}\) (luôn đúng)
Vậy điều cần chứng minh là đúng
2) \(\sqrt{x^2-5x+4}+2\sqrt{x+5}=2\sqrt{x-4}+\sqrt{x^2+4x-5}\)
⇔ \(\sqrt{\left(x-4\right)\left(x-1\right)}-2\sqrt{x-4}+2\sqrt{x+5}-\sqrt{\left(x+5\right)\left(x-1\right)}=0\)
⇔ \(\sqrt{x-4}.\left(\sqrt{x-1}-2\right)-\sqrt{x+5}\left(\sqrt{x-1}-2\right)=0\)
⇔ \(\left(\sqrt{x-4}-\sqrt{x+5}\right)\left(\sqrt{x-1}-2\right)=0\)
⇔ \(\left[{}\begin{matrix}\sqrt{x-4}-\sqrt{x+5}=0\\\sqrt{x-1}-2=0\end{matrix}\right.\)
⇔ \(\left[{}\begin{matrix}\sqrt{x-4}=\sqrt{x+5}\\\sqrt{x-1}=2\end{matrix}\right.\)
⇔ \(\left[{}\begin{matrix}x\in\varnothing\\x=5\end{matrix}\right.\)
⇔ x = 5
Vậy S = {5}
1) \(\sqrt[]{9\left(x-1\right)}=21\)
\(\Leftrightarrow9\left(x-1\right)=21^2\)
\(\Leftrightarrow9\left(x-1\right)=441\)
\(\Leftrightarrow x-1=49\Leftrightarrow x=50\)
2) \(\sqrt[]{1-x}+\sqrt[]{4-4x}-\dfrac{1}{3}\sqrt[]{16-16x}+5=0\)
\(\Leftrightarrow\sqrt[]{1-x}+\sqrt[]{4\left(1-x\right)}-\dfrac{1}{3}\sqrt[]{16\left(1-x\right)}+5=0\)
\(\)\(\Leftrightarrow\sqrt[]{1-x}+2\sqrt[]{1-x}-\dfrac{4}{3}\sqrt[]{1-x}+5=0\)
\(\Leftrightarrow\sqrt[]{1-x}\left(1+3-\dfrac{4}{3}\right)+5=0\)
\(\Leftrightarrow\sqrt[]{1-x}.\dfrac{8}{3}=-5\)
\(\Leftrightarrow\sqrt[]{1-x}=-\dfrac{15}{8}\)
mà \(\sqrt[]{1-x}\ge0\)
\(\Leftrightarrow pt.vô.nghiệm\)
3) \(\sqrt[]{2x}-\sqrt[]{50}=0\)
\(\Leftrightarrow\sqrt[]{2x}=\sqrt[]{50}\)
\(\Leftrightarrow2x=50\Leftrightarrow x=25\)
1) \(\sqrt{9\left(x-1\right)}=21\) (ĐK: \(x\ge1\))
\(\Leftrightarrow3\sqrt{x-1}=21\)
\(\Leftrightarrow\sqrt{x-1}=7\)
\(\Leftrightarrow x-1=49\)
\(\Leftrightarrow x=49+1\)
\(\Leftrightarrow x=50\left(tm\right)\)
2) \(\sqrt{1-x}+\sqrt{4-4x}-\dfrac{1}{3}\sqrt{16-16x}+5=0\) (ĐK: \(x\le1\))
\(\Leftrightarrow\sqrt{1-x}+2\sqrt{1-x}-\dfrac{4}{3}\sqrt{1-x}+5=0\)
\(\Leftrightarrow\dfrac{5}{3}\sqrt{1-x}+5=0\)
\(\Leftrightarrow\dfrac{5}{3}\sqrt{1-x}=-5\) (vô lý)
Phương trình vô nghiệm
3) \(\sqrt{2x}-\sqrt{50}=0\) (ĐK: \(x\ge0\))
\(\Leftrightarrow\sqrt{2x}=\sqrt{50}\)
\(\Leftrightarrow2x=50\)
\(\Leftrightarrow x=\dfrac{50}{2}\)
\(\Leftrightarrow x=25\left(tm\right)\)
4) \(\sqrt{4x^2+4x+1}=6\)
\(\Leftrightarrow\sqrt{\left(2x+1\right)^2}=6\)
\(\Leftrightarrow\left|2x+1\right|=6\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+1=6\left(ĐK:x\ge-\dfrac{1}{2}\right)\\2x+1=-6\left(ĐK:x< -\dfrac{1}{2}\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=5\\2x=-7\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\left(tm\right)\\x=-\dfrac{7}{2}\left(tm\right)\end{matrix}\right.\)
5) \(\sqrt{\left(x-3\right)^2}=3-x\)
\(\Leftrightarrow\left|x-3\right|=3-x\)
\(\Leftrightarrow x-3=3-x\)
\(\Leftrightarrow x+x=3+3\)
\(\Leftrightarrow x=\dfrac{6}{2}\)
\(\Leftrightarrow x=3\)
\(\sqrt{4x^2-4x+1}=3-x\left(x\in R\right)\\ \Leftrightarrow\sqrt{\left(2x-1\right)^2}=3-x\\ \Leftrightarrow2x-1=3-x\\ \Leftrightarrow3x=4\Leftrightarrow x=\dfrac{4}{3}\\ \sqrt{9x+9}+\sqrt{x+1}-\sqrt{4x+4}=2\left(x+1\right)\left(x\ge-1\right)\\ \Leftrightarrow\sqrt{x+1}\left(\sqrt{9}+1+\sqrt{4}\right)=2\left(x+1\right)\\ \Leftrightarrow6\sqrt{x+1}=2\left(x+1\right)\\ \Leftrightarrow3\sqrt{x+1}=x+1\\ \Leftrightarrow\sqrt{x+1}\left(3-\sqrt{x+1}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x+1=0\\\sqrt{x+1}=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x+1=9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\left(tm\right)\\x=8\left(tm\right)\end{matrix}\right.\)
a, ĐK: \(x\in R\)
\(\sqrt{4x^2-4x+1}=3-x\)
\(\Leftrightarrow\sqrt{\left(2x-1\right)^2}=3-x\)
\(\Leftrightarrow\left|2x-1\right|=3-x\)
TH1: \(\left\{{}\begin{matrix}2x-1\ge0\\2x-1=3-x\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{1}{2}\\x=\dfrac{4}{3}\end{matrix}\right.\Leftrightarrow x=\dfrac{4}{3}\)
TH2: \(\left\{{}\begin{matrix}2x-1< 0\\1-2x=3-x\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x< \dfrac{1}{2}\\x=-2\end{matrix}\right.\Leftrightarrow x=-2\)
Thấy : \(x^2-4x+16=\left(x-2\right)^2+12>0\forall x\)
P/t \(\Leftrightarrow2\left(x^2-4x+16\right)-36+\sqrt{x^2-4x+16}=0\)
Đặt \(t=\sqrt{x^2-4x+16}>0\) ; khi đó :
\(2t^2+t-36=0\) \(\Leftrightarrow\left[{}\begin{matrix}t=4\\t=-\dfrac{9}{2}\left(L\right)\end{matrix}\right.\)
Với t = 4 hay \(\sqrt{x^2-4x+16}=4\Leftrightarrow x^2-4x=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)
Vậy ...
a: Ta có: \(\sqrt{1-x^2}=x-1\)
\(\Leftrightarrow1-x^2=x-1\)
\(\Leftrightarrow1-x^2-x+1=0\)
\(\Leftrightarrow x^2+x-2=0\)
\(\Leftrightarrow\left(x+2\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\left(loại\right)\\x=1\left(nhận\right)\end{matrix}\right.\)
b: Ta có: \(\sqrt{x^2+4x+4}=x-2\)
\(\Leftrightarrow\left|x+2\right|=x-2\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=x-2\left(x\ge-2\right)\\x+2=2-x\left(x< -2\right)\end{matrix}\right.\Leftrightarrow2x=0\)
hay x=0(loại)
Do \(x^6-x^3+x^2-x+1=\left(x^3-\dfrac{1}{2}\right)^2+\left(x-\dfrac{1}{2}\right)^2+\dfrac{1}{2}>0\) ; \(\forall x\) nên BPT tương đương:
\(\sqrt{13}-\sqrt{2x^2-2x+5}-\sqrt{2x^2-4x+4}\ge0\)
\(\Leftrightarrow\sqrt{4x^2-4x+10}+\sqrt{4x^2-8x+8}\le\sqrt{26}\) (1)
Ta có:
\(VT=\sqrt{\left(2x-1\right)^2+3^2}+\sqrt{\left(2-2x\right)^2+2^2}\ge\sqrt{\left(2x-1+2-2x\right)^2+\left(3+2\right)^2}=\sqrt{26}\) (2)
\(\Rightarrow\left(1\right);\left(2\right)\Rightarrow\sqrt{4x^2-4x+10}+\sqrt{4x^2-8x+8}=\sqrt{26}\)
Dấu "=" xảy ra khi và chỉ khi \(2\left(2x-1\right)=3\left(2-2x\right)\Leftrightarrow x=\dfrac{4}{5}\)
Vậy BPT có nghiệm duy nhất \(x=\dfrac{4}{5}\)