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ĐK: \(x^3+3x^2-3x+1\ge0\)
\(pt\Leftrightarrow\sqrt[3]{9x^2-15x+9}-\left(2-x\right)+\sqrt{x^3+3x^2-3x+1}=0\)
\(\Leftrightarrow\frac{9x^2-15x+9-\left(2-x\right)^3}{A^2+AB+B^2}+\sqrt{x^3+3x^2-3x+1}=0\)
\(\left(A=\sqrt[3]{9x^2-15x+9};\text{ }B=2-x\right)\)\(\text{(}A^2+AB+B^2=\left(A+\frac{B}{2}\right)^2+\frac{3B^2}{4}>0\text{)}\)
\(\Leftrightarrow\frac{x^3+3x^2-3x+1}{A^2+AB+B^2}+\sqrt{x^3+3x^2-3x+1}=0\)
\(\Leftrightarrow\sqrt{x^3+3x^2-3x+1}\left(\frac{\sqrt{x^3+3x^2-3x+1}}{A^2+AB+B^2}+1\right)=0\)
\(\Leftrightarrow x^3+3x^2-3x+1=0\text{ (do }\frac{\sqrt{x^3+3x^2-3x+1}}{A^2+AB+B^2}+1>0\text{)}\)
\(\Leftrightarrow\left(x+1+\sqrt[3]{2}+\sqrt[3]{4}\right)\left[x^2+\left(2-\sqrt[3]{2}-\sqrt[3]{4}\right)x+\sqrt[3]{2}-1\right]=0\)
\(\Leftrightarrow x+1+\sqrt[3]{2}+\sqrt[3]{4}=0\text{ (}pt\text{ }x^2+\left(2-\sqrt[3]{2}-\sqrt[3]{4}\right)x+\sqrt[3]{2}-1=0\text{ vô nghiệm do }\Delta
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ĐKXĐ:...
\(\sqrt{3x^2-5x-1}-\sqrt{3x^2-7x+9}+\sqrt{x^2-2}-\sqrt{x^2-3x+13}=0\)
\(\Leftrightarrow\frac{2\left(x-5\right)}{\sqrt{3x^2-5x-1}+\sqrt{3x^2-7x+9}}+\frac{3\left(x-5\right)}{\sqrt{x^2-2}+\sqrt{x^2-3x+13}}=0\)
\(\Leftrightarrow\left(x-5\right)\left(\frac{2}{\sqrt{3x^2-5x-1}+\sqrt{3x^2-7x+9}}+\frac{3}{\sqrt{x^2-2}+\sqrt{x^2-3x+13}}\right)=0\)
\(\Leftrightarrow x-5=0\) (ngoặc to phía sau luôn dương)
\(\Rightarrow x=5\)
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vô đây Câu hỏi của Phan hữu Dũng - Toán lớp 9 - Học toán với OnlineMath
Cho mình copy nhé:
Đặt \(\sqrt{3x-2}=a;\sqrt{x-1}=b\left(a,b\ge0\right)\)
\(\Rightarrow\begin{cases}a^2=3x-2\\b^2=x-1\end{cases}\)\(\Rightarrow a^2+b^2=4x-3\)
\(pt\Leftrightarrow a+b=a^2+b^2-6+2ab\)
\(\Leftrightarrow a^2+b^2-6+2ab-a-b=0\)
\(\Leftrightarrow\left(a+b\right)^2-\left(a+b\right)-6=0\)
\(\Leftrightarrow\left(a+b\right)^2+2\left(a+b\right)-3\left(a+b\right)-6=0\)
\(\Leftrightarrow\left(a+b\right)\left(a+b+2\right)-3\left(a+b+2\right)=0\)
\(\Leftrightarrow\left(a+b-3\right)\left(a+b+2\right)=0\)
\(\Leftrightarrow a+b=3\)hoặc\(a+b=-2\)(loại,vì a\(\ge\)0;b\(\ge\)0 =>a+b\(\ge\)0)
- Với a+b=3
\(\Rightarrow\sqrt{3x-2}+\sqrt{x-1}=3\)
\(\Leftrightarrow\sqrt{3x-2}=3-\sqrt{x-1}\)
\(\Rightarrow3x-2=9+x-1-6\sqrt{x-1}\)
\(\Rightarrow2x-10=-6\sqrt{x-1}\)
\(\Rightarrow4x^2-40x+100=36\left(x-1\right)\)
\(\Rightarrow4x^2-76x+1236=0\)
\(\Rightarrow4x^2-8x-68x+136=0\)
\(\Rightarrow4x\left(x-2\right)-68\left(x-2\right)=0\)
\(\Rightarrow\left(4x-68\right)\left(x-2\right)=0\)
\(\Rightarrow\left[\begin{array}{nghiempt}x=17\left(loai\right)\\x=2\left(TM\right)\end{array}\right.\)
Vậy phương trình đã cho có nghiệm là x=2
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a.-3x^2+15x=0\)
\(\Leftrightarrow3x\left(-x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x=0\\-x+5=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\end{matrix}\right.\)
\(b.2x^2-32=0\)
\(\Leftrightarrow2x^2=32\)
\(\Leftrightarrow x^2=16\)
\(\Leftrightarrow\left|x\right|=4\)
\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-4\end{matrix}\right.\)
\(c.2x^2-5x+1=0\)
\(a=2;b=-5;c=1\)
\(\Delta=\left(-5\right)^2-4.2.1=17>0\)
Do \(\Delta>0\) nên phương trình có 2 nghiệm phân biệt:
\(x_1=\dfrac{5+\sqrt{17}}{4}\)
\(x_2=\dfrac{5-\sqrt{17}}{4}\)
\(a,-3x^2+15x=0\\ -3x\left(x-5\right)=0\\ \Rightarrow\left\{{}\begin{matrix}x=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=5\end{matrix}\right.\)
\(b,\\ 2\left(x^2-16\right)=0\\ \Leftrightarrow x^2-16=0\\ \Leftrightarrow\left(x-4\right)\left(x+4\right)=0\\ \Rightarrow\left\{{}\begin{matrix}x=4\\x=-4\end{matrix}\right.\)
\(c,\\ \Delta=5^2-4.2=17\\ \Rightarrow x_1,x_2=\dfrac{\Delta\pm b}{2ac}\\ =\dfrac{5\pm\sqrt{17}}{4}\)
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\(\Leftrightarrow\sqrt{12-7x}-\sqrt{x^2-x}=\sqrt{3x^2-5x-1}-\sqrt{x^2-3x+4}\)
\(\Rightarrow-\sqrt{3x^2-5x-1}-\sqrt{x^2-x}+\sqrt{x^2-3x+4}+\sqrt{12-7x}=0\)
=>\(x\approx-3,4579061804411\)
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