Question

bài 3: Tìm x, biết:

a)(2x-1)²-25=0               b)8x³-50x=0

c)2(x+3)-x²-3x=0      d)x³+27+(x+3)(x-9)=0

mình cần gấppppppppp

Answer 1

a) ${}^{}-25=0$

${}^{}=0+25=25$

${}^{}={}^{}={}^{}$

$\Rightarrow [\begin{array}{c}2x-1=5\\ 2x-1=-5\end{array}\Rightarrow [\begin{array}{c}2x=6\\ 2x=-4\end{array}\Rightarrow [\begin{array}{c}x=3\\ x=-2\end{array}$

b) $8{}^{}-50x=0$

$2x(4{}^{}$

Answer 2

câu b thiếu bn ơi

Answer 3

ok

a) ${}^{}-25=0$

${}^{}=0+25=25$

${}^{}={}^{}={}^{}$

$\Rightarrow [\begin{array}{c}2x-1=5\\ 2x-1=-5\end{array}\Rightarrow [\begin{array}{c}2x=6\\ 2x=-4\end{array}\Rightarrow [\begin{array}{c}x=3\\ x=-2\end{array}$

b) $8{}^{}-50x=0$

$2x(4{}^{}$

Answer 4

b) $8{}^{}-50x=0$

$2x\left(4{}^{}-25\right)=0$

$\Rightarrow [\begin{array}{c}2x=0\\ 4{}^{}-25=0\end{array}\Rightarrow ⎡⎢\; ⎢⎣\begin{array}{c}x=0\\ 4{}^{}=25\Rightarrow {}^{}=\frac{25}{4}\Rightarrow [\begin{array}{c}x=\frac{5}{2}\\ x=-\frac{5}{2}\end{array}\end{array}$

Answer 5

hoặc tham thảo nhé

Answer 6

a. (2x - 1)² - 25 = 0

<=> (2x - 1 - 25)(2x - 1 + 25) = 0

<=> (2x - 26)(2x + 24) = 0

<=> \(\left[{}\begin{matrix}2x-26=0\\2x+24=0\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}x=13\\x=-12\end{matrix}\right.\)ư

b. 8x³ - 50x = 0

<=> 2x(4x² - 25) = 0

<=> 2x(2x - 5)(2x + 5) = 0

<=> \(\left[{}\begin{matrix}2x=0\\2x-5=0\\2x+5=0\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}x=0\\x=\dfrac{5}{2}\\x=\dfrac{-5}{2}\end{matrix}\right.\)

c. 2(x + 3) - x² - 3x = 0

<=> 2(x + 3) - x(x + 3) = 0

<=> (2 - x)(x + 3) = 0

<=> \(\left[{}\begin{matrix}2-x=0\\x+3=0\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)

d. x³ + 27 + (x + 3)(x - 9) = 0

<=> (x + 3)(x² - 3x + 9) + (x + 3)(x - 9) = 0

<=> (x + 3)(x² - 3x + 9 + x - 9) = 0

<=> (x + 3)(x² - 2x) = 0

<=> x(x + 3)(x - 2) = 0

<=> \(\left[{}\begin{matrix}x=0\\x+3=0\\x-2=0\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}x=0\\x=-3\\x=2\end{matrix}\right.\)

Answer 7

nó bị lỗi đấy cậu

Answer 8

Tiếp câu a:

\(\left(2x-26\right)\left(2x+24\right)=0\)

<=> \(\left[{}\begin{matrix}2x-26=0\\2x+24=0\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}x=13\\x=-12\end{matrix}\right.\)

Answer 9

Tương tự vậy luôn nhé