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Nó có 1 nghiệm là 9
Bạn chứng minh nó là nghiệm duy nhất đi
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\(\left(\sqrt{2x^2-15x+26}\right)\)^2=\(\left(x-4\right)^2\) (ĐKXĐ;x>=4)
\(2x^2\)-15x+26=\(^{x^2-8x+16}\)
\(x^2\)-7x+10=0
\(x^2\)-7x+\(\frac{49}{4}\)=\(\frac{9}{4}\)
(x-\(\frac{7}{4}\))^2=\(\left(\frac{3}{2}\right)^2\)
x-\(\frac{7}{4}\) =\(\frac{3}{2}\)
x=3.25 (khong tm dk)
v...................
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2x3 - 15x2 + 26x - 5 = 0
<=> 2x3 - 10x2 - 5x2 + 25x + x - 5 = 0
<=> 2x2( x - 5 ) - 5x( x - 5 ) + ( x - 5 ) = 0
<=> ( x - 5 )( 2x2 - 5x + 1 ) = 0
<=> \(\orbr{\begin{cases}x-5=0\\2x^2-5x+1=0\end{cases}}\)
+) x - 5 = 0 <=> x = 5
+) 2x2 - 5x + 1 = 0
Δ = b2 - 4ac = (-5)2 - 4.2.1 = 25 - 8 = 17
Δ > 0, áp dụng công thức nghiệm thu được \(x_1=\frac{5+\sqrt{17}}{4};x_2=\frac{5-\sqrt{17}}{4}\)
Vậy phương trình đã cho có ba nghiệm \(x_1=\frac{5+\sqrt{17}}{4};x_2=\frac{5-\sqrt{17}}{4};x_3=5\)
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\(\left(2x^2+3\right)^2-10x^2-15x=0\)
\(\Leftrightarrow4x^4+12x^2+9-10x^2-15x=0\)
\(\Leftrightarrow4x^4+2x^2-15x+9=0\)
\(\Leftrightarrow4x^4-4x^2+6x^2-6x-9x+9=0\)
\(\Leftrightarrow4x^2\left(x^2-1\right)+6x\left(x-1\right)-9\left(x-1\right)=0\)
\(\Leftrightarrow4x^2\left(x-1\right)\left(x+1\right)+6x\left(x-1\right)-9\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left[4x\left(x+1\right)+6x-9\right]=0\)
\(\Leftrightarrow\left(x-1\right)\left(4x^2+10x-9\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(4x^2+10x+\frac{25}{4}+\frac{11}{4}\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left[\left(2x+\frac{5}{2}\right)^2+\frac{11}{4}\right]=0\)
Vì \(\left(2x+\frac{5}{2}\right)^2+\frac{11}{4}>0\)
=> x - 1 = 0
=> x = 1
Vậy x = 1
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\(15x^4+30x^3+13x^2-2x-1=0\)
<=> \(15x^4+15x^3+15x^3+15x^2-2x^2-2x-1=0\)
<=> \(15x^2\left(x^2+x\right)+15x\left(x^2+x\right)-2\left(x^2+x\right)-1\)
<=> \(15\left(x^2+x\right)^2-2\left(x^2+x\right)-1=0\)
<=> \(\orbr{\begin{cases}x^2+x=\frac{1}{3}\\x^2+x=\frac{1}{5}\end{cases}}\)
Em tự giải tiếp nhé!
\(-2x^2+15x-5=\sqrt{2x^2-15x+11}\)
\(pt\Leftrightarrow-2x^2+15x-7=\sqrt{2x^2-15x+11}-2\)
\(\Leftrightarrow-\left(x-7\right)\left(2x-1\right)=\dfrac{2x^2-15x+11-4}{\sqrt{2x^2-15x+11}+2}\)
\(\Leftrightarrow-\left(x-7\right)\left(2x-1\right)-\dfrac{\left(x-7\right)\left(2x-1\right)}{\sqrt{2x^2-15x+11}+2}=0\)
\(\Leftrightarrow-\left(x-7\right)\left(2x-1\right)\left(1+\dfrac{1}{\sqrt{2x^2-15x+11}+2}\right)=0\)
Dễ thấy:\(1+\dfrac{1}{\sqrt{2x^2-15x+11}+2}>0\)
\(\Rightarrow\left[{}\begin{matrix}x-7=0\\2x-1=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=7\\x=\dfrac{1}{2}\end{matrix}\right.\)