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ĐKXĐ: \(2\le x\le5\)
\(\left(\sqrt{2x-4}-\sqrt{5-x}\right)\sqrt{3x-3}=3x-9\)
\(\Leftrightarrow\dfrac{\left(3x-9\right)\sqrt{3x-3}}{\sqrt{2x-4}+\sqrt{5-x}}=3x-9\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-9=0\Rightarrow x=3\\\dfrac{\sqrt{3x-3}}{\sqrt{2x-4}+\sqrt{5-x}}=1\left(1\right)\end{matrix}\right.\)
Xét (1):
\(\Leftrightarrow\sqrt{3x-3}=\sqrt{2x-4}+\sqrt{5-x}\)
\(\Leftrightarrow3x-3=x+1+2\sqrt{\left(2x-4\right)\left(5-x\right)}\)
\(\Leftrightarrow x-2=\sqrt{\left(2x-4\right)\left(5-x\right)}\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge2\\\left(x-2\right)^2=\left(2x-4\right)\left(5-x\right)\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge2\\\left(x-2\right)\left(3x-12\right)=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2\\x=4\end{matrix}\right.\)
Vậy pt có 3 nghiệm \(x=\left\{2;3;4\right\}\)
\(\hept{\begin{cases}x^2\left(y+3\right)\left(x-2\right)-\sqrt{2x+3}=0\left(1\right)\\4x-4\sqrt{\left(2x+3\right)}+x^3\sqrt{\left(y+3\right)^2}+9=0\left(2\right)\end{cases}}\)
Ta có:
\(\left(2\right)\Leftrightarrow x^2|y+3|=\frac{4\sqrt{2x+3}-4x-9}{x}\)
\(\Leftrightarrow\orbr{\begin{cases}x^2\left(y+3\right)=\frac{4\sqrt{2x+3}-4x-9}{x}\left(3\right)\\x^2\left(y+3\right)=-\frac{4\sqrt{2x+3}-4x-9}{x}\left(4\right)\end{cases}}\)
Thế (3) vô (1) được
\(\frac{4\sqrt{2x+3}-4x-9}{x}.\left(x-2\right)-\sqrt{2x+3}=0\)
Đặt \(\hept{\begin{cases}\sqrt{2x+3}=a\ge0\\x=\frac{a^2-3}{2}\end{cases}}\)
\(\Rightarrow\left(4a-2\left(a^2-3\right)-9\right)\left(\frac{a^2-3}{2}-2\right)-a\left(\frac{a^2-3}{2}\right)=0\)
Làm đến đây thì thấy nó phương trình bậc 4 thôi bỏ. Phương trình bậc 4 giải tốn công. Xem như 1 hướng đi.
Xem lại đề là \(\left(x-2\right)\)hay \(\left(x+2\right)\)nhé. Nghiệm xấu quá.
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) \(\sqrt{\left(x-3\right)^2}=3-x\)
\(\Leftrightarrow\left|x-3\right|=3-x\)
\(\Leftrightarrow x-3=3-x\)
\(\Leftrightarrow x+x=3+3\)
\(\Leftrightarrow x=\dfrac{6}{2}\)
\(\Leftrightarrow x=3\)
\(BPT\Leftrightarrow x\sqrt[3]{25x\left(2x^2+9\right)}\le4x^2+3\\ \Leftrightarrow\sqrt[3]{25x^4\left(2x^2+9\right)}\le4x^2+3\left(1\right)\)
Áp dụng BĐT cosi:
\(\sqrt[3]{5x^2\cdot5x^2\left(2x^2+9\right)}\le\dfrac{5x^2+5x^2+2x^2+9}{3}=\dfrac{12x^2+9}{3}=4x^2+3\)
Vậy \(\left(1\right)\) luôn đúng
Dấu \("="\Leftrightarrow5x^2=2x^2+9\Leftrightarrow x^2=3\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{3}\\x=-\sqrt{3}\end{matrix}\right.\)
Tham khảo:
1) Giải phương trình : \(11\sqrt{5-x}+8\sqrt{2x-1}=24+3\sqrt{\left(5-x\right)\left(2x-1\right)}\) - Hoc24
=căn(căn x+9)^2 -3^2=24
= x+9-9=2x
=0=2x-x
X=0