\(x^2-5x-36=0\)

\(\Leftrightarrow x^2+4x-9x-36=0\)

\(\Leftrightarrow x\left(x+4\right)-9\left(x+4\right)=0\)

\(\Leftrightarrow\left(x-9\right)\left(x+4\right)=0\Leftrightarrow x=-4;x=9\)

\(x^2-5x-36=0\)

\(x^2+4x-9x-36=0\)

\(x\left(x-9\right)+4\left(x-9\right)=0\)

\(\left(x-9\right)\left(x+4\right)=0\)

\(\orbr{\begin{cases}x-9=0\\x+4=0\end{cases}\orbr{\begin{cases}x=9\left(TM\right)\\x=-4\left(TM\right)\end{cases}}}\)

a, \(x^2-2.\frac{1}{3}x+\frac{1}{9}=\left(x-\frac{1}{3}\right)^2\)

Thay x = 9 vào ta được : \(=\left(9-\frac{1}{3}\right)^2=\left(\frac{26}{3}\right)^2=\frac{676}{9}\)

\(x^2-\frac{2}{3}x+\frac{1}{9}\)

Thay \(x=9\) và ta được:

\(9^2-\frac{2}{3}9+\frac{1}{9}\)\(=81-6+\frac{1}{9}\)\(=\frac{676}{9}\)

Trả lời:

1, x³ - x² - 4

= x³ - x² - 4 + 2x - 2x

= x³ - 2x² + x² - 4 + 2x - 2x

= ( x³ + x² + 2x ) - ( 2x² + 2x + 4 )

= x ( x² + x + 2 ) - 2 ( x² + x + 2 )

= ( x - 2 )( x² + x + 2 )

2, x³ - 2x - 4 

= x³ - 2x - 4 + 2x² - 2x²

= x³ - 4x + 2x - 4 + 2x² - 2x²

= ( x³ + 2x² + 2x ) - ( 2x² + 4x + 4 )

= x ( x² + 2x + 2 ) - 2 ( x² + 2x + 2 )

= ( x - 2 )( x² + 2x + 2 )

giúp mình đi gấp lm rùi

(dak bủh bủn lmao lmao ) đoạn này dành cho Đào Phúc Khánh

a,676 phần 9

b,Ko bt,sao lại có y?

đm con chó

e, \(\left(x^3-4x^2\right)-\left(x-4\right)=0\)

\(\Leftrightarrow x^2\left(x-4\right)-\left(x-4\right)=0\Leftrightarrow\left(x^2-1\right)\left(x-4\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(x-1\right)\left(x-4\right)=0\Leftrightarrow x=\pm1;x=4\)

f, \(2x^3-242x=0\Leftrightarrow2x\left(x^2-121\right)=0\)

\(\Leftrightarrow2x\left(x-11\right)\left(x+11\right)=0\Leftrightarrow x=\pm11;x=0\)

g, \(x^5-9x=0\Leftrightarrow x\left(x^4-9\right)=0\)

\(\Leftrightarrow x\left(x^2-3\right)\left(x^2+3>0\right)=0\Leftrightarrow x=\pm\sqrt{3};x=0\)

a, \(5x\left(x-1\right)+\left(x+17\right)=0\)

\(\Leftrightarrow5x^2-5x+x+17=0\Leftrightarrow5x^2-4x+17=0\)

\(\Leftrightarrow5\left(x^2-\frac{4}{5}x\right)+17=0\Leftrightarrow5\left(x^2-2.\frac{2}{5}x+\frac{4}{25}-\frac{4}{25}\right)+17=0\)

\(\Leftrightarrow5\left(x-\frac{2}{5}\right)^2-\frac{4}{5}+17=0\Leftrightarrow5\left(x-\frac{2}{5}\right)^2+81\ge81>0\)

Vậy pt vô nghiệm 

b, \(3x\left(x-3\right)^2-3x\left(x+3\right)^2=0\)

\(\Leftrightarrow3x\left[\left(x-3\right)^2-\left(x+3\right)^2\right]=0\)

\(\Leftrightarrow3x\left(x-3-x-3\right)\left(x-3+x+3\right)=0\Leftrightarrow x.2x=0\Leftrightarrow x=0\)

c, \(2x^2-9x+7=0\Leftrightarrow2x^2-7x-2x+7=0\)

\(\Leftrightarrow x\left(2x-7\right)-\left(2x-7\right)=0\Leftrightarrow\left(x-1\right)\left(2x-7\right)=0\Leftrightarrow x=1;x=\frac{7}{2}\)

Trả lời:

a, \(5x\left(x-1\right)+\left(x+17\right)=0\)

\(\Leftrightarrow5x^2-5x+x+17=0\)

\(\Leftrightarrow5x^2-4x+17=0\)

\(\Leftrightarrow5\left(x^2-\frac{4}{5}x+\frac{17}{5}\right)=0\)

\(\Leftrightarrow x^2-\frac{4}{5}x+\frac{17}{5}=0\)

\(\Leftrightarrow x^2-2.x.\frac{2}{5}+\frac{4}{25}+\frac{81}{25}=0\)

\(\Leftrightarrow\left(x-\frac{2}{5}\right)^2+\frac{81}{25}=0\)

Vì \(\left(x-\frac{2}{5}\right)^2+\frac{81}{25}\ge\frac{81}{25}>0\forall x\)

nên pt vô nghiệm 

b, \(3x\left(x-3\right)^2-3x\left(x+3\right)^2=0\)

\(\Leftrightarrow3x\left[\left(x-3\right)^2-\left(x+3\right)^2\right]=0\)

\(\Leftrightarrow3x\left(x-3-x-3\right)\left(x-3+x+3\right)=0\)

\(\Leftrightarrow3x.\left(-9\right).2x=0\)

\(\Leftrightarrow-54x^2=0\)

\(\Leftrightarrow x^2=0\)

\(\Leftrightarrow x=0\)

Vậy x = 0 là nghiệm của pt.

c, \(7-9x+2x^2=0\)

\(\Leftrightarrow2x^2-7x-2x+7=0\)

\(\Leftrightarrow x\left(2x-7\right)-\left(2x-7\right)=0\)

\(\Leftrightarrow\left(2x-7\right)\left(x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}2x-7=0\\x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{7}{2}\\x=1\end{cases}}}\)

Vậy x = 7/2; x = 1 là nghiệm của pt.

d, trùng ý c

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