giair phương trình \(x^2+6x+1=\left(2x+1\right)\sqrt{x^2+2x+3}\)
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25 tháng 1 2021
a) Ta có: \(2x^3+5x^2-3x=0\)
\(\Leftrightarrow x\left(2x^2+5x-3\right)=0\)
\(\Leftrightarrow x\left(2x^2+6x-x-3\right)=0\)
\(\Leftrightarrow x\left[2x\left(x+3\right)-\left(x+3\right)\right]=0\)
\(\Leftrightarrow x\left(x+3\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+3=0\\2x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\\2x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\\x=\dfrac{1}{2}\end{matrix}\right.\)
Vậy: \(S=\left\{0;-3;\dfrac{1}{2}\right\}\)
b) Ta có: \(2x^3+6x^2=x^2+3x\)
\(\Leftrightarrow2x^2\left(x+3\right)=x\left(x+3\right)\)
\(\Leftrightarrow2x^2\left(x+3\right)-x\left(x+3\right)=0\)
\(\Leftrightarrow x\left(x+3\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+3=0\\2x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\\2x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\\x=\dfrac{1}{2}\end{matrix}\right.\)
Vậy: \(S=\left\{0;-3;\dfrac{1}{2}\right\}\)
c) Ta có: \(x^2+\left(x+2\right)\left(11x-7\right)=4\)
\(\Leftrightarrow x^2+11x^2-7x+22x-14-4=0\)
\(\Leftrightarrow12x^2+15x-18=0\)
\(\Leftrightarrow12x^2+24x-9x-18=0\)
\(\Leftrightarrow12x\left(x+2\right)-9\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(12x-9\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\12x-9=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\12x=9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{3}{4}\end{matrix}\right.\)
Vậy: \(S=\left\{-2;\dfrac{3}{4}\right\}\)
25 tháng 1 2021
a) Ta có: \(\left(x-\sqrt{2}\right)+3\left(x^2-2\right)=0\)
\(\Leftrightarrow\left(x-\sqrt{2}\right)+3\left(x-\sqrt{2}\right)\left(x+\sqrt{2}\right)=0\)
\(\Leftrightarrow\left(x-\sqrt{2}\right)\left(1+3x+3\sqrt{2}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\sqrt{2}=0\\3x+3\sqrt{2}+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{2}\\3x=-3\sqrt{2}-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{2}\\x=\dfrac{-3\sqrt{2}-1}{3}\end{matrix}\right.\)
Vậy: \(S=\left\{\sqrt{2};\dfrac{-3\sqrt{2}-1}{3}\right\}\)
b) Ta có: \(x^2-5=\left(2x-\sqrt{5}\right)\left(x+\sqrt{5}\right)\)
\(\Leftrightarrow\left(x+\sqrt{5}\right)\left(x-\sqrt{5}\right)-\left(2x-\sqrt{5}\right)\left(x+\sqrt{5}\right)=0\)
\(\Leftrightarrow\left(x+\sqrt{5}\right)\left(x-\sqrt{5}-2x+\sqrt{5}\right)=0\)
\(\Leftrightarrow-x\left(x+\sqrt{5}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}-x=0\\x+\sqrt{5}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\sqrt{5}\end{matrix}\right.\)
Vậy: \(S=\left\{0;-\sqrt{5}\right\}\)
KC
1
19 tháng 6 2021
ta có:
pt trên \(< =>x^2+6x+1=\left(2x+1\right)\sqrt{x^2+2x+3}\)
\(< =>\left[\left(x^2+6x\right)+1\right]^2=\left(2x+1\right)^2.\left(x^2+2x+3\right)\)
\(< =>x^4+12x^3+36x^2+2.\left(x^2+6x\right)+1=\left(4x^2+4x+1\right)\left(x^2+2x+3\right)\)
\(< =>x^4+12x^3+38x^2+12x+1=\)
\(4x^4+8x^3+12x^2+4x^3+8x^2+12x+x^2+2x+3\)
\(=4x^4+12x^3+21x^2+14x+3\)
\(< =>-3x^4+17x^2-2x-2=0\)
\(< =>-\left(x^2+2x-1\right)\left(3x^2-6x+2\right)=0\)
đến đây dễ rùi bạn tự giải nhé
BC
0
BC
1
NA
20 tháng 5 2022
\(\text{Đ}K:x^2+2x+3\ge0\\ x^2+6x+1=\left(2x+1\right)\cdot\sqrt{x^2+2x+3}\\ \Leftrightarrow x^2+2x+3+4x+2=\left(2x+1\right)\cdot\sqrt{x^2+2x+3+4}\)
\(\text{ Đặt }\)\(m=\sqrt{x^2+2x+3};n=2x+1\) \(\text{ phương trình trở thành :}\)
\(m^2+2n=mn+4\\ \Leftrightarrow m^2-4-mn+2n=0\\ \Leftrightarrow\left(m-2\right)\left(m+2\right)-n\left(m-2\right)=0\\ \Leftrightarrow\left(m-2\right)\left(m-n-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}m=2\\m-n=-2\end{matrix}\right.\)
`\text{ Với}` \(m=2\\ \Leftrightarrow\sqrt{x^2+2x+3}=2\Leftrightarrow x^2+2x-1=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\sqrt{2}-1\left(N\right)\\x=-\sqrt{2}-1\left(N\right)\end{matrix}\right.\)
`\text{Với}`\(m-n=-2\Leftrightarrow\sqrt{x^2+2x+3}-\left(2x+1\right)=-2\\ \Leftrightarrow\sqrt{x^2+2x+3}=-2+2x+1=2x-1\\ \Leftrightarrow x^2+2x+3=4x^2-4x+1\\ \Leftrightarrow3x^2-6x-2=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3+\sqrt{15}}{3}\left(N\right)\\x=\dfrac{3-\sqrt{15}}{3}\left(L\right)\end{matrix}\right.\)
ĐK: \(x^2+2x+3>0\)(BĐT đúng) (Tự Cm được)
Với đk trên, đặt:
\(\hept{\begin{cases}\sqrt{x^2+2x+3}=a\\2x+1=b\end{cases}}\)với a > 0
\(\Leftrightarrow\hept{\begin{cases}a^2=x^2+2x+3\\2b=4x+2\end{cases}\Rightarrow a^2+2b=x^2+6x+5}\)
Pt trở thành
\(a^2+2b-4=ab\)
\(\Leftrightarrow4a^2+8b-16=4ab\)
\(\Leftrightarrow4a^2-4ab=-8b+16\)
\(\Leftrightarrow4a^2-4ab+b^2=b^2-8b+16\)
\(\Leftrightarrow\left(2a-b\right)^2=\left(b-4\right)^2\)
Đến đây tự làm nha