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- Tại câu a có hai bạn ra hai đáp án nên góp thêm lựa ý kiến cho người hỏi đỡ bối rối

a, Ta có : \(\left(x-3\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+9\left(x+1\right)^2=15\)

=> \(\left(x-3\right)\left(x^2-6x+9-x^2-3x-9\right)+9\left(x+1\right)^2=15\)

=> ​​​\(9\left(x+1\right)^2-9\left(x^2-3x\right)=15\)​

=> \(9\left(x^2+2x+1-x^2+3x\right)=15\)

=> \(9\left(5x+1\right)=15\)

=> \(5x=\frac{15}{9}-1=\frac{2}{3}\)

=> \(x=\frac{2}{15}\)

Vậy ....

a/ \(25x^2-9=0\)

<=> \(\left(5x-3\right)\left(5x+3\right)=0\)

<=> \(\orbr{\begin{cases}5x-3=0\\5x+3=0\end{cases}}\)

<=> \(\orbr{\begin{cases}5x=3\\5x=-3\end{cases}}\)

<=> \(\orbr{\begin{cases}x=\frac{3}{5}\\x=-\frac{3}{5}\end{cases}}\)

b/ \(\left(x+4\right)^2-\left(x+9\right)\left(x-1\right)=16\)

<=> \(x^2+8x+16-x^2+8x-9=16\)

<=> \(16x+7=16\)

<=> \(16x=9\)

<=> \(x=\frac{9}{16}\)

a) \(25x^2-9=0\)

\(\Leftrightarrow\left(5x-3\right)\left(5x+3\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}5x-3=0\\5x+3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}5x=3\\5x=-3\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{3}{5}\\x=-\frac{3}{5}\end{cases}}}\)

Vậy S = {3/5 ; -3/5}

b) \(\left(x+4\right)^2-\left(x+9\right)\left(x-1\right)=16\)

\(\Leftrightarrow\left(x+4\right)^2-4^2-\left(x+9\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left(x+4-4\right)\left(x+4+4\right)-\left(x+9\right)\left(x-1\right)=0\)

\(\Leftrightarrow x\left(x+8\right)-\left(x+9\right)\left(x-1\right)=0\)

\(\Leftrightarrow x^2+8x-x^2-8x+9=0\)

\(\Leftrightarrow9=0\left(vl\right)\)

Vậy S = \(\varnothing\)

Giải như sau.

(1)+(2)⇔x2−2x+1+√x2−2x+5=y2+√y2+4⇔(x2−2x+5)+√x2−2x+5=y2+4+√y2+4⇔√y2+4=√x2−2x+5⇒x=3y(1)+(2)⇔x2−2x+1+x2−2x+5=y2+y2+4⇔(x2−2x+5)+x2−2x+5=y2+4+y2+4⇔y2+4=x2−2x+5⇒x=3y

⇔√y2+4=√x2−2x+5⇔y2+4=x2−2x+5, chỗ này do hàm số f(x)=t2+tf(x)=t2+t đồng biến ∀t≥0∀t≥0
Công việc còn lại là của bạn ! 

\(\left(x+6\right)\left(2x+1\right)=0\)

<=>  \(\orbr{\begin{cases}x+6=0\\2x+1=0\end{cases}}\)

<=>  \(\orbr{\begin{cases}x=-6\\x=-\frac{1}{2}\end{cases}}\)

Vậy....

hk tốt

^^

a) x³ - 16x = 0

x(x² - 16) = 0

=> x = 0 hoặc x² - 16 = 0

x = 4

Vậy x = 0 hoặc x = 4

b) x⁴ -2x³ + 10x² - 20x = 0

x³ (x - 2) + 10x(x - 2) = 0

(x - 2)(x³ + 10x) = 0

=> x - 2 = 0 hoặc x³ + 10x = 0

x = 2 x(x² + 10) = 0

+ TH1: x = 0

+ TH2: x² + 10 = 0

x² = -10 (vô lí)

Vậy x = 2 hoặc x = 0

c) (2x - 3)² = (x + 5)²

(2x)² + 2 . 2x . 3 + 3² = x² + 2.x.5 + 5²

4x² + 12x + 9 = x² + 10x + 25

4x² + 12x - x² - 10x = 25 - 9

3x² + 2x = 16

x(3x + 2) = 16

Đến đây bạn làm nốt câu c nhé!

a) \(4x^3-9x=0\)

\(\Leftrightarrow x\left(4x^2-9\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\4x^2=9\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm\frac{3}{2}\end{cases}}\)

b) \(3x\left(x-2\right)-5x+10=0\)

\(\Leftrightarrow\left(3x-5\right)\left(x-2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{5}{3}\\x=2\end{cases}}\)

c) \(4x\left(x+3\right)-x^2+9=0\)

\(\Leftrightarrow4x\left(x+3\right)-\left(x-3\right)\left(x+3\right)=0\)

\(\Leftrightarrow\left(3x+3\right)\left(x+3\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=-1\\x=-3\end{cases}}\)

d) \(\left(2x+5\right)\left(x-4\right)=\left(x-4\right)\left(5-x\right)\)

\(\Leftrightarrow\left(2x+5\right)\left(x-4\right)+\left(x-5\right)\left(x-4\right)=0\)

\(\Leftrightarrow3x\left(x-4\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=4\end{cases}}\)

e) \(16x^2-25=\left(4x-5\right)\left(2x+1\right)\)

\(\Leftrightarrow\left(4x-5\right)\left(4x+5\right)-\left(4x-5\right)\left(2x+1\right)=0\)

\(\Leftrightarrow\left(4x-5\right)\left(2x+4\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{5}{4}\\x=-2\end{cases}}\)

f) \(\left(x+\frac{1}{5}\right)^2=\frac{64}{9}\)

\(\Leftrightarrow\orbr{\begin{cases}x+\frac{1}{5}=\frac{8}{3}\\x+\frac{1}{5}=-\frac{8}{3}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{37}{15}\\x=-\frac{43}{15}\end{cases}}\)

g) \(9\left(x+2\right)^2=\left(x+3\right)^2\)

\(\Leftrightarrow\orbr{\begin{cases}3x+6=x+3\\3x+6=-x-3\end{cases}}\Leftrightarrow\orbr{\begin{cases}2x=-3\\4x=-9\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-\frac{3}{2}\\x=-\frac{9}{4}\end{cases}}\)