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Nguyễn Hà Giang
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mit béo
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Nguyễn Lê Phước Thịnh
19 tháng 12 2021 lúc 11:54

g: \(\Leftrightarrow\left(x^2+6x+5\right)\left(x^2+6x+8\right)-4=0\)

\(\Leftrightarrow\left(x^2+6x\right)^2+13\left(x^2+6x\right)+36=0\)

\(\Leftrightarrow\left(x+3\right)^2\left(x^2+6x+4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=\sqrt{5}-3\\x=-\sqrt{5}-3\end{matrix}\right.\)

Tùng
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Akai Haruma
6 tháng 1 2024 lúc 12:13

Lời giải:
a. $(x^2-9)(5x+15)=0$

$\Rightarrow x^2-9=0$ hoặc $5x+15=0$
Nếu $x^2-9=0$

$\Rightarrow x^2=9=3^2=(-3)^2$

$\Rightarrow x=3$ hoặc $-3$
Nếu $5x+15=0$

$\Rightarrow x=-3$
b.

$x^2-8x=0$
$\Rightarrow x(x-8)=0$

$\Rightarrow x=0$ hoặc $x-8=0$

$\Rightarrow x=0$ hoặc $x=8$

c. 

$5+12(x-1)^2=53$

$12(x-1)^2=53-5=48$

$(x-1)^2=48:12=4=2^2=(-2)^2$

$\Rightarrow x-1=2$ hoặc $x-2=-2$
$\Rightarrow x=3$ hoặc $x=0$

d.

$(x-5)^2=36=6^2=(-6)^2$
$\Rightarrow x-5=6$ hoặc $x-5=-6$

$\Rightarrow x=11$ hoặc $x=-1$

e.

$(3x-5)^3=64=4^3$

$\Rightarrow 3x-5=4$

$\Rightarrow 3x=9$

$\Rightarrow x=3$

f.

$4^{2x}+2^{4x+3}=144$
$2^{4x}+2^{4x}.8=144$

$2^{4x}(1+8)=144$

$2^{4x}.9=144$

$2^{4x}=144:9=16=2^4$

$\Rightarrow 4x=4\Rightarrow x=1$

Trần Ngọc Mỹ
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Monkey D. Luffy
17 tháng 11 2021 lúc 9:59

\(1,\Leftrightarrow x\left(x-9\right)=0\Leftrightarrow\left[{}\begin{matrix}x=9\\x=0\end{matrix}\right.\\ 2,\Leftrightarrow x^2-4x-x^2=7\Leftrightarrow-4x=7\Leftrightarrow x=-\dfrac{7}{4}\\ 3,\Leftrightarrow3x+2x-10=5\Leftrightarrow5x=15\Leftrightarrow x=3\\ 4,\Leftrightarrow\left(5x-1\right)\left(5x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=-\dfrac{1}{5}\end{matrix}\right.\\ 5,\Leftrightarrow\left(x-2\right)\left(3x-5\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{5}{3}\end{matrix}\right.\\ 6,\Leftrightarrow\left(x-7\right)\left(3x+4\right)=0\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-\dfrac{4}{3}\end{matrix}\right.\)

\(7,\Leftrightarrow\left(2x-3\right)\left(2x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\\ 8,\Leftrightarrow\left(x-4\right)\left(10x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{5}\\x=4\end{matrix}\right.\\ 9,\Leftrightarrow2x^2-5x-2x^2=0\Leftrightarrow x=0\\ 10,\Leftrightarrow2x\left(x-2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\\ 11,\Leftrightarrow\left(4x-3\right)\left(3-2x\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=\dfrac{3}{2}\end{matrix}\right.\\ 12,\Leftrightarrow2x^2-10x-2x^2=3\Leftrightarrow-10x=3\Leftrightarrow x=-\dfrac{3}{10}\)

ILoveMath
17 tháng 11 2021 lúc 10:00

\(1,\Leftrightarrow x\left(x-9\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=9\end{matrix}\right.\\ 2,\Leftrightarrow x^2-4x-x^2=7\\ \Leftrightarrow-4x=7\\ \Leftrightarrow x=\dfrac{-7}{4}\\ 3,\Leftrightarrow3x+2x-10=5\\ \Leftrightarrow5x=15\\ \Leftrightarrow x=3\\ 4,\Leftrightarrow\left(5x-1\right)\left(5x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=-\dfrac{1}{5}\end{matrix}\right.\)

\(5,\Leftrightarrow\left(x-2\right)\left(3x-5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{5}{3}\end{matrix}\right.\\ 6,\Leftrightarrow\left(3x+4\right)\left(x-7\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{4}{3}\\x=7\end{matrix}\right.\\ 7,\Leftrightarrow\left(2x-3\right)\left(2x+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)

\(8,\Leftrightarrow10x\left(x-4\right)+2\left(x-4\right)=0\\ \Leftrightarrow\left(x-4\right)\left(10x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=4\\x=-\dfrac{1}{5}\end{matrix}\right.\\ 9,\Leftrightarrow2x^2-5x-2x^2=0\\ \Leftrightarrow-5x=0\\ \Leftrightarrow x=0\\ 10,\Leftrightarrow2x\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)

\(11,\Leftrightarrow\left(2x-3\right)\left(4x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{3}{4}\end{matrix}\right.\\ 12,\Leftrightarrow2x^2-10x-2x^2=3\\ \Leftrightarrow-10x=3\\ \Leftrightarrow x=-\dfrac{3}{10}\)

i love Vietnam
17 tháng 11 2021 lúc 10:11

1) \(x^2-9x=0\Rightarrow x\left(x-9\right)=0\Rightarrow x=0;9\)

2) \(x\left(x-4\right)-x^2=7\Rightarrow-4x=7\Rightarrow x=-\dfrac{7}{4}\)

3) \(3x+2\left(x-5\right)=5\Rightarrow5x-10=5\Rightarrow5x=15\Rightarrow x=3\)

4) \(25x^2-1=0\Rightarrow x^2=\dfrac{1}{25}\Rightarrow x=\pm\dfrac{1}{5}\)

5) \(3x\left(x-2\right)-5\left(x-2\right)=0\Rightarrow\left(x-2\right)\left(3x-5\right)=0\Rightarrow x=2;\dfrac{5}{3}\)

6) \(3x\left(x-7\right)+4\left(x-7\right)\Rightarrow\left(3x+4\right)\left(x-7\right)=0\Rightarrow x=-\dfrac{4}{3};7\)

7) \(4x^2-9=0\Rightarrow x^2=\dfrac{9}{4}\Rightarrow x=\pm\dfrac{3}{2}\)

8) \(10x\left(x-4\right)+2x-8=0\Rightarrow2\left(x-4\right)\left(5x+1\right)=0\Rightarrow x=4;-\dfrac{1}{5}\)

9) \(x\left(2x-5\right)-2x^2=0\Rightarrow x\left(2x-5-2x=0\right)\Rightarrow x=0\)

10) \(2x^2-4x=0\Rightarrow2x\left(x-2\right)=0\Rightarrow x=0;2\)

11) \(2x\left(3-4x\right)+3\left(4x-3\right)=0\Rightarrow2x\left(4x-3\right)-3\left(4x-3\right)=0\Rightarrow\left(4x-3\right)\left(2x-3\right)=0\Rightarrow x=\dfrac{3}{4};\dfrac{3}{2}\)

12) \(2x\left(x-5\right)-2x^2=3\Rightarrow-10x=3\Rightarrow x=-\dfrac{3}{10}\)

nood
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Ngô Hải Nam
11 tháng 7 2023 lúc 21:41

a)

\(x^2-4\sqrt{15}x+19=0\\ < =>x^2-4\sqrt{15}x+60-41=0\\ < =>\left(x-2\sqrt{15}\right)^2-41=0\\ < =>\left(x-2\sqrt{15}-\sqrt{41}\right)\left(x-2\sqrt{15}+\sqrt{41}\right)=0\\ < =>\left[{}\begin{matrix}x-2\sqrt{15}-\sqrt{41}=0\\x-2\sqrt{15}+\sqrt{41}=0\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=2\sqrt{15}+\sqrt{41}\\x=2\sqrt{15}-\sqrt{41}\end{matrix}\right.\)

b)

\(4x^2+4\sqrt{5}x+5=0\\ < =>\left(2x+\sqrt{5}\right)^2=0\\ < =>2x+\sqrt{5}=0\\ < =>2x=-\sqrt{5}\\ < =>-\dfrac{\sqrt{5}}{2}\)

Nguyễn Lê Phước Thịnh
11 tháng 7 2023 lúc 21:35

a: Δ=(4căn 15)^2-4*1*19=164>0

Phương trình có hai nghiệm phân biệt là:

\(\left\{{}\begin{matrix}x=\dfrac{4\sqrt{5}-2\sqrt{41}}{2}=2\sqrt{5}-\sqrt{41}\\x_2=2\sqrt{5}+\sqrt{41}\end{matrix}\right.\)

b: \(\Leftrightarrow\left(2x\right)^2+2\cdot2x\cdot\sqrt{5}+5=0\)

=>(2x+căn 5)^2=0

=>2x+căn 5=0

=>x=-1/2*căn 5

Ngọc Khánh
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Nguyễn Hoàng Minh
13 tháng 11 2021 lúc 17:30

\(1,\Leftrightarrow x^2+10x+25=x^2-4x-21\\ \Leftrightarrow14x=-46\\ \Leftrightarrow x=-\dfrac{23}{7}\\ 2,\Leftrightarrow x^3+8=15+x^3+2x\\ \Leftrightarrow2x=-7\Leftrightarrow x=-\dfrac{7}{2}\\ 3,\Leftrightarrow\left(x+3\right)^2=0\\ \Leftrightarrow x=-3\\ 4,\Leftrightarrow x^3-9x^2+27x-27=0\\ \Leftrightarrow\left(x-3\right)^3=0\\ \Leftrightarrow x-3=0\Leftrightarrow x=3\\ 5,\Leftrightarrow4x^2+4x+1-4x^2-16x-16=9\\ \Leftrightarrow-12x=24\Leftrightarrow x=-2\\ 6,\Leftrightarrow x^2-3x+5x-15=0\\ \Leftrightarrow\left(x-3\right)\left(x+5\right)=0\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-5\end{matrix}\right.\)

Nguyễn An Bình
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Nguyễn thành Đạt
29 tháng 1 2023 lúc 15:38

\(\dfrac{4}{3}\times\dfrac{7}{5}\times2:\dfrac{34}{9}\times\dfrac{15}{4}\)

\(=\left(\dfrac{4}{3}\times\dfrac{15}{4}\right)\times\left(\dfrac{14}{5}\times\dfrac{9}{34}\right)\)

\(=5\times\dfrac{63}{85}\)

\(=\dfrac{63}{17}\)

Trần Nguyễn Xuân Khôi
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Nguyễn Minh Anh
7 tháng 1 2022 lúc 20:35

3) \(-12+2x-9+x=0\\ -21+3x=0\\ 3x=21\\ x=7\)

Vân Vui Vẻ
7 tháng 1 2022 lúc 20:42

4)
\(11+\left(15-x\right)=1\)
\(15-x=1-11\)
\(15-x=-10\)
\(x=15-\left(-10\right)\)
\(x=25\)

5) 
\(4-\left(27-3\right)=x-\left(13-4\right)\)
\(4-24=x-9\)
\(x-9=-20\)
\(x=-20+9\)
\(x=-11\)
 

Thanh Hoàng Thanh
7 tháng 1 2022 lúc 20:47

\(3.-12+\left(2x-9\right)+x=0.\)

\(\Leftrightarrow-12+2x-9+x=0.\Leftrightarrow3x=21.\Leftrightarrow x=7.\)

Vậy \(x=7.\)

\(4.11+\left(15-x\right)=1.\Leftrightarrow11+15-x=1.\Leftrightarrow26-x=1.\Leftrightarrow x=25.\)

Vậy \(x=25.\)

\(5.4-\left(27-3\right)=x-\left(13-4\right).\Leftrightarrow4-24=x-9.\Leftrightarrow-20=x-9.\Leftrightarrow x=-11.\)

Vậy \(x=-11.\)

\(6.8-\left(x-10\right)=23-\left(-4+12\right).\Leftrightarrow8-x+10=23-8.\Leftrightarrow18-x=15.\Leftrightarrow x=3.\)

Vậy \(x=3.\)

\(7.105-5\left(10-5x\right)=-20.\Leftrightarrow105-50+25x=-20.\Leftrightarrow25x=-75.\Leftrightarrow x=-3.\)

Vậy \(x=-3.\)

\(8.\left(x-1\right)\left(8-2x\right)\left(3x+123\right)=0.\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0.\\8-2x=0.\\3x+123=0.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1.\\x=4.\\x=-41.\end{matrix}\right.\)

Vậy \(x\in\left\{1;4;-41\right\}.\)

\(9.\left(x^2-25\right)\left(x+10\right)=0.\)

\(\Leftrightarrow\left(x-5\right)\left(x+5\right)\left(x+10\right)=0.\)

\(\Leftrightarrow\left[{}\begin{matrix}x-5=0.\\x+5=0.\\x+10=0.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5.\\x=-5.\\x=-10.\end{matrix}\right.\)

Vậy \(x\in\left\{5;-5;-10\right\}.\)

\(10.x\left(x^2+5\right)=0.\Leftrightarrow x=0.\)

Pham Trong Bach
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Cao Minh Tâm
24 tháng 6 2019 lúc 5:29

a) Rút gọn VT = 45x + 8. Từ đó tìm được x = 2 15 .  

b) Rút gọn VT = -25x – 8. Từ đó tìm được x = − 11 25 .