Giải phương trình: \(\sqrt{2x-1}+x^2-3x+1=0\)
Giải các phương trình, bất phương trình sau:
1) \(\sqrt{3x+7}-5< 0\)
2) \(\sqrt{-2x-1}-3>0\)
3) \(\dfrac{\sqrt{3x-2}}{6}-3=0\)
4) \(-5\sqrt{-x-2}-1< 0\)
5) \(-\dfrac{2}{3}\sqrt{-3-x}-3>0\)
1) \(\sqrt[]{3x+7}-5< 0\)
\(\Leftrightarrow\sqrt[]{3x+7}< 5\)
\(\Leftrightarrow3x+7\ge0\cap3x+7< 25\)
\(\Leftrightarrow x\ge-\dfrac{7}{3}\cap x< 6\)
\(\Leftrightarrow-\dfrac{7}{3}\le x< 6\)
Giải phương trình:
a) \(\sqrt{3x+1}-\sqrt{6-x}+3x^2-14x-8=0\)
b) \(\sqrt{2x^2-1}+x\sqrt{2x-1}=2x^2\)
c) \(\dfrac{2\sqrt{2}}{\sqrt{x+1}}+\sqrt{x}=\sqrt{x+9}\)
b)đk:\(x\ge\dfrac{1}{2}\)
Có: \(\sqrt{2x^2-1}\le\dfrac{2x^2-1+1}{2}=x^2\)
\(x\sqrt{2x-1}=\sqrt{\left(2x^2-x\right)x}\le\dfrac{2x^2-x+x}{2}=x^2\)
=>\(\sqrt{2x^2-1}+x\sqrt{2x-1}\le2x^2\)
Dấu = xảy ra\(\Leftrightarrow x=1\)
Vậy....
c) đk: \(x\ge0\)
\(\Leftrightarrow\sqrt{x}=\sqrt{x+9}-\dfrac{2\sqrt{2}}{\sqrt{x+1}}\)
\(\Rightarrow x=x+9+\dfrac{8}{x+1}-4\sqrt{\dfrac{2\left(x+9\right)}{x+1}}\)
\(\Leftrightarrow0=9+\dfrac{8}{x+1}-4\sqrt{\dfrac{2\left(x+9\right)}{x+1}}\)
Đặt \(a=\sqrt{\dfrac{2\left(x+9\right)}{x+1}}\left(a>0\right)\)
\(\Leftrightarrow\dfrac{a^2-2}{2}=\dfrac{8}{x+1}\)
pttt \(9+\dfrac{a^2-2}{2}-4a=0\) \(\Leftrightarrow a=4\) (TM)
\(\Rightarrow4=\sqrt{\dfrac{2\left(x+9\right)}{x+1}}\) \(\Leftrightarrow16=\dfrac{2\left(x+9\right)}{x+1}\) \(\Leftrightarrow x=\dfrac{1}{7}\) (TM)
Vậy ...
a)ĐKXĐ: x≥-1/3; x≤6
<=>\(\dfrac{3x-15}{\sqrt{3x+1}+4}+\dfrac{x-5}{\sqrt{x-6}+1}+\left(x-5\right)\cdot\left(3x+1\right)=0\Leftrightarrow\left(x-5\right)\cdot\left(\dfrac{3}{\sqrt{3x+1}+4}+\dfrac{1}{\sqrt{x-6}+1}+3x+1\right)=0\Leftrightarrow x-5=0\Leftrightarrow x=5\)(nhận)
(vì x≥-1/3 nên3x+1≥0 )
Giải phương trình
\(-3x^2+x+3+\left(\sqrt{3x+2}-4\right)\sqrt{3x-2x^2}+\left(x-1\right)\sqrt{3x+2}=0\)
Giải các phương trình sau
\(1)\sqrt{x}+\sqrt{x^2-1}=\sqrt{2x^2-3x-4}\)
\(2)x^3+\left(3x^2-4x-4\right)\sqrt{x+1}=0\)
1.
ĐKXĐ: \(x\ge\dfrac{3+\sqrt{41}}{4}\)
\(\Leftrightarrow x^2+x-1+2\sqrt{x\left(x^2-1\right)}=2x^2-3x-4\)
\(\Leftrightarrow x^2-4x-3-2\sqrt{\left(x^2-x\right)\left(x+1\right)}=0\)
Đặt \(\left\{{}\begin{matrix}\sqrt{x^2-x}=a>0\\\sqrt{x+1}=b>0\end{matrix}\right.\)
\(\Rightarrow a^2-3b^2-2ab=0\)
\(\Leftrightarrow\left(a+b\right)\left(a-3b\right)=0\)
\(\Leftrightarrow a=3b\)
\(\Leftrightarrow\sqrt{x^2-x}=3\sqrt{x+1}\)
\(\Leftrightarrow x^2-x=9\left(x+1\right)\)
\(\Leftrightarrow...\) (bạn tự hoàn thành nhé)
2.
ĐKXĐ: \(x\ge-1\)
Đặt \(\sqrt{x+1}=a\ge0\) pt trở thành:
\(x^3+3\left(x^2-4a^2\right)a=0\)
\(\Leftrightarrow x^3+3ax^2-4a^3=0\)
\(\Leftrightarrow\left(x-a\right)\left(x+2a\right)^2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=x\\2a=-x\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+1}=x\left(x\ge0\right)\\2\sqrt{x+1}=-x\left(x\le0\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2=x+1\\x^2=4x+4\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-x-1=0\\x^2-4x-4=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1+\sqrt{5}}{2}\\x=2-2\sqrt{2}\end{matrix}\right.\)
Giải phương trình sau \(\sqrt{2x-1}+x^2-3x+1=0\)
\(ĐK:x\ge\dfrac{1}{2}\\ PT\Leftrightarrow\sqrt{2x-1}=-x^2+3x-1\\ \Leftrightarrow2x-1=\left(-x^2+3x-1\right)^2=\left(x^2-3x+1\right)^2\\ \Leftrightarrow2x-1=x^4+9x^2+1-6x^3-6x+2x^2\\ \Leftrightarrow x^4-6x^3+11x^2-8x+2=0\\ \Leftrightarrow x^4-x^3-5x^3+5x^2+6x^2-6x-2x+2=0\\ \Leftrightarrow\left(x-1\right)\left(x^3-5x^2+6x-2\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x^3-x^2-4x^2+4x+2x-2\right)=0\\ \Leftrightarrow\left(x-1\right)^2\left(x^2-4x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\left(tm\right)\\x^2-4x+2=0\left(1\right)\end{matrix}\right.\\ \Delta\left(1\right)=16-8=8\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4-2\sqrt{2}}{2}=2-\sqrt{2}\left(tm\right)\\x=\dfrac{4+2\sqrt{2}}{2}=2+\sqrt{2}\left(tm\right)\end{matrix}\right.\\ S=\left\{1;2-\sqrt{2};2+\sqrt{2}\right\}\)
Giải các phương trình sau:
1) \(\sqrt{3x^2+5x+8}-\sqrt{3x^2+5x+1}=1\)
2) \(x^2-2x-12+4\sqrt{\left(4-x\right)\left(2+x\right)}=0\)
3) \(3\sqrt{x}+\dfrac{3}{2\sqrt{x}}=2x+\dfrac{1}{2x}-7\)
4) \(\sqrt{x}-\dfrac{4}{\sqrt{x+2}}+\sqrt{x+2}=0\)
5)\(\left(x-7\right)\sqrt{\dfrac{x+3}{x-7}}=x+4\)
6) \(2\sqrt{x-4}+\sqrt{x-1}=\sqrt{2x-3}+\sqrt{4x-16}\)
7) \(\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}=\dfrac{x+3}{2}\)
Giúp mình với ajk, mink đang cần gấp
Giải phương trình \(\sqrt{3x+1}+\sqrt{2x-1}+x^2+2x-6=0\) bằng phương pháp đặt ẩn phụ
1. giải phương trình chứa căn bậc 2
a) \(\sqrt{x^2-x+1}=x\)
b) \(\sqrt{x^2-3x+2}+\sqrt{x^2+x-6}=0\)
c) \(\sqrt{x^4-2x^2+1}=x-1\)
a.
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x^2-x+1=x^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\1-x=0\end{matrix}\right.\)
\(\Leftrightarrow x=1\)
b.
ĐKXĐ: \(\left[{}\begin{matrix}x\ge2\\x\le-3\end{matrix}\right.\)
Do \(\left\{{}\begin{matrix}\sqrt{x^2-3x+2}\ge0\\\sqrt{x^2+x-6}\ge0\end{matrix}\right.\) với mọi x thuộc TXĐ
\(\Rightarrow\sqrt{x^2-3x+2}+\sqrt{x^2+x-6}\ge0\)
Đẳng thức xảy ra khi:
\(\left\{{}\begin{matrix}x^2-3x+2=0\\x^2+x-6=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\\\left\{{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow x=2\) (thỏa mãn ĐKXĐ)
Vậy pt có nghiệm duy nhất \(x=2\)
c.
Với \(x< 1\Rightarrow\left\{{}\begin{matrix}x-1< 0\\\sqrt{x^4-2x^2+1}\ge0\end{matrix}\right.\) phương trình vô nghiệm
Với \(x\ge1\) pt tương đương:
\(\sqrt{\left(x^2-1\right)^2}=x-1\)
\(\Leftrightarrow\left|x^2-1\right|=x-1\)
\(\Leftrightarrow x^2-1=x-1\) (do \(x\ge1\Rightarrow x^2-1\ge0\Rightarrow\left|x^2-1\right|=x-1\))
\(\Leftrightarrow x^2-x=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0< 1\left(loại\right)\\x=1\end{matrix}\right.\)
Giải phương trình sau
1. \(5x^2-16x+7+\left(x+1\right)\sqrt{x^2+3x-1}=0\)
2. \(3\left(\sqrt{2x^2+1}-1\right)=x\left(1+3x+8\sqrt{2x^2+1}\right)\)
\(\left(\frac{2x-1}{2-x}+2\sqrt{2-x}\right)^3=27\left(2x-1\right)\)
Giải phương trình nghiệm nguyên sau:
\(3x^3-13x^2+30x-4=\sqrt{\left(6x+2\right)\left(3x-4\right)^3}\)
Giải phương trình và hệ phương trình:
1) \(-2x^2+x+1-2\sqrt{x^2+x+1}=0\)
2) \(\left\{{}\begin{matrix}x^4+y^3x+x^2y^2=3y^4\\2x^2+y^4+1=2x\left(y^2+1\right)\end{matrix}\right.\)
1) \(-2x^2+x+1-2\sqrt[]{x^2+x+1}=0\)
\(\Leftrightarrow2\sqrt[]{x^2+x+1}=-2x^2+x+1\left(1\right)\)
Ta có :
\(2\sqrt[]{x^2+x+1}=2\sqrt[]{\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}}\ge\sqrt[]{3}\)
Dấu "=" xảy ra khi và chỉ khi \(x+\dfrac{1}{2}=0\Leftrightarrow x=-\dfrac{1}{2}\)
\(\left(1\right)\Leftrightarrow-2x^2+x+1=\sqrt[]{3}\)
\(\Leftrightarrow2x^2-x+\sqrt[]{3}-1=0\)
\(\Delta=1-8\left(\sqrt[]{3}-1\right)=9-8\sqrt[]{3}\)
\(pt\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1+\sqrt[]{9-8\sqrt[]{3}}}{4}\left(loại\right)\\x=\dfrac{1-\sqrt[]{9-8\sqrt[]{3}}}{4}\left(loại\right)\end{matrix}\right.\) \(\left(vì.x=-\dfrac{1}{2}\right)\)
Vậy phương trình cho vô nghiệm