Bài 1: Tìm x
a)\(4x^2-12x=9\)
b)\(3.\left(x^2-4\right)-5x.\left(x+2\right)=0\)
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LM
0
30 tháng 7 2021
1)\(\sqrt{4x^2+12x+9}=2-x\)
\(\Leftrightarrow\sqrt{\left(2x+3\right)^2}=2-x\)
\(\Leftrightarrow\left|2x+3\right|=2-x\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+3=2-x\\2x+3=x-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=-1\\x=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{3}\\x=-5\end{matrix}\right.\)
\(\)
17 tháng 12 2023
a: ĐKXĐ: \(x\in R\)
\(\sqrt{\left(2x+3\right)^2}=5\)
=>|2x+3|=5
=>\(\left[{}\begin{matrix}2x+3=5\\2x+3=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=2\\2x=-8\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=1\left(nhận\right)\\x=-4\left(nhận\right)\end{matrix}\right.\)
b: ĐKXĐ: \(x\in R\)
\(\sqrt{9\left(x-2\right)^2}=18\)
=>\(\sqrt{9}\cdot\sqrt{\left(x-2\right)^2}=18\)
=>\(3\cdot\left|x-2\right|=18\)
=>\(\left|x-2\right|=6\)
=>\(\left[{}\begin{matrix}x-2=6\\x-2=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\left(nhận\right)\\x=-4\left(nhận\right)\end{matrix}\right.\)
c: ĐKXĐ: x>=2
\(\sqrt{9x-18}-\sqrt{4x-8}+3\sqrt{x-2}=40\)
=>\(3\sqrt{x-2}-2\sqrt{x-2}+3\sqrt{x-2}=40\)
=>\(4\sqrt{x-2}=40\)
=>\(\sqrt{x-2}=10\)
=>x-2=100
=>x=102(nhận)
d: ĐKXĐ: \(x\in R\)
\(\sqrt{4\left(x-3\right)^2}=8\)
=>\(\sqrt{\left(2x-6\right)^2}=8\)
=>|2x-6|=8
=>\(\left[{}\begin{matrix}2x-6=8\\2x-6=-8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=14\\2x=-2\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=7\left(nhận\right)\\x=-1\left(nhận\right)\end{matrix}\right.\)
e: ĐKXĐ: \(x\in R\)
\(\sqrt{4x^2+12x+9}=5\)
=>\(\sqrt{\left(2x\right)^2+2\cdot2x\cdot3+3^2}=5\)
=>\(\sqrt{\left(2x+3\right)^2}=5\)
=>|2x+3|=5
=>\(\left[{}\begin{matrix}2x+3=5\\2x+3=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=2\\2x=-8\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=1\left(nhận\right)\\x=-4\left(nhận\right)\end{matrix}\right.\)
f: ĐKXĐ:x>=6/5
\(\sqrt{5x-6}-3=0\)
=>\(\sqrt{5x-6}=3\)
=>\(5x-6=3^2=9\)
=>5x=6+9=15
=>x=15/5=3(nhận)
16 tháng 12 2022
1: \(\Leftrightarrow2x^2-10x-3x-2x^2=0\)
=>-13x=0
=>x=0
2: \(\Leftrightarrow5x-2x^2+2x^2-2x=13\)
=>3x=13
=>x=13/3
3: \(\Leftrightarrow4x^4-6x^3-4x^3+6x^3-2x^2=0\)
=>-2x^2=0
=>x=0
4: \(\Leftrightarrow5x^2-5x-5x^2+7x-10x+14=6\)
=>-8x=6-14=-8
=>x=1
2
16 tháng 12 2022
`1)2x(x-5)-(3x+2x^2)=0`
`<=>2x^2-10x-3x-2x^2=0`
`<=>-13x=0`
`<=>x=0`
___________________________________________________
`2)x(5-2x)+2x(x-1)=13`
`<=>5x-2x^2+2x^2-2x=13`
`<=>3x=13<=>x=13/3`
___________________________________________________
`3)2x^3(2x-3)-x^2(4x^2-6x+2)=0`
`<=>4x^4-6x^3-4x^4+6x^3-2x^2=0`
`<=>x=0`
___________________________________________________
`4)5x(x-1)-(x+2)(5x-7)=0`
`<=>5x^2-5x-5x^2+7x-10x+14=0`
`<=>-8x=-14`
`<=>x=7/4`
___________________________________________________
`5)6x^2-(2x-3)(3x+2)=1`
`<=>6x^2-6x^2-4x+9x+6=1`
`<=>5x=-5<=>x=-1`
___________________________________________________
`6)2x(1-x)+5=9-2x^2`
`<=>2x-2x^2+5=9-2x^2`
`<=>2x=4<=>x=2`
KM
6 tháng 4 2020
câu a, b, c dễ mà. Bạn áp dụng 7 hằng đẳng thúc là làm đc thoii!!
vd: a) \(\left(9x^2-4\right)\left(x+1\right)=\left(3x+2\right)\left(x^2-1\right)\)
\(\Rightarrow\left(3x-2\right)\left(3x+2\right)\left(x+1\right)=\left(3x+2\right)\left(x-1\right)\left(x+1\right)\)
\(\Rightarrow\left(3x-2\right)\left(3x+2\right)-\left(3x+2\right)\left(x-1\right)\left(x+1\right)=0\)
\(\Rightarrow\left(3x+2\right)\left(x+1\right)[\left(3x-2\right)-\left(x-1\right)]=0\)
\(\Rightarrow\left(3x+2\right)\left(x+1\right)\left(2x-1\right)=0\) (bạn phá ngoặc ra rồi tính là ra bước này)
\(\Leftrightarrow3x+2=0\) hoặc \(x+1=0\) hoặc \(2x-1=0\) ( đến đây bạn chia làm 3 trường hợp r tự tính nhé)
Chúc bạn học tốt!!
NV
Nguyễn Việt Lâm
Giáo viên
6 tháng 4 2020
d/
\(\Leftrightarrow x^3\left(x+1\right)+\left(x+1\right)=0\)
\(\Leftrightarrow\left(x^3+1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\x^3+1=0\end{matrix}\right.\) \(\Rightarrow x=-1\)
e/
\(\Leftrightarrow x^3+x^2-6x-x^2-x+6=0\)
\(\Leftrightarrow x\left(x^2+x-6\right)-\left(x^2+x-6\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2+x-6\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-2\right)\left(x+3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=2\\x=-3\end{matrix}\right.\)
31 tháng 7 2018
1,x^2-(x+1)(x-1)=0
x^2-x^2+1+0
1=0(vô lý)
2,5x^3+3x^2+3x+1=4x^2
x^3+3x^2+3x+1=0
(x+1)=0
x=-1
3,x^3+x^2=0
x^2(x+1)=0
x=0 or x=-1
4,2x^3-12x^2+18x=0
x^3-6x^2+9x=0
x(x^2-6x+9)=0
x(x-3)^2=0
x=0 or x=3
5,5x^2-4(x^2-2x+1)+20=0
5x^2-4x^2+8x-4+20=0
x^2+8x+16=0
(x+4)^2=0
x=-4
6,5x(x-3)+7x-21=0
5x(x-3)+7(x-3)=0
(5x+7)(x-3)=0
5x-7=0 or x-3=0
x=7/5 or x=3
7,2x^3-50x=0
2x(x^2-25)=0
2x(x-5)(x+5)=0
x=0 or x=5 or x=-5
8,(4x-1)^2-9(x+3)^2=0
(4x-1)^2-3^2*(x+3)^2=0
(4x-1)^2-(3x+9)^2=0
(4x-1-3x-9)(4x-1+3x+9)=0
(x-10)(7x+8)=0
x=10 or x=-8/7
9,3(x-2)^2-x+2=0
3*(x-2)*(x-2)-(x-2)=0
(3x-6)(x-2)-(x-2)=0
(x-2)(3x-6-1)=0
(x-2)(3x-7)=0
x=2 or x=7/3
10,9x^2+6x-8=0
9x^2+12x-6x-8=0
3x(3x-2)+4(3x-2)=0
(3x+4)(3x-2)=0
3x+4=0 or 3x-2=0
x=-4/3 or x=2/3
23 tháng 10 2016
a) \(4x^2-12x=-9\)
\(\Leftrightarrow4x^2-12x+9=0\)
\(\Leftrightarrow\left(2x-3\right)^2=0\)
\(\Leftrightarrow2x-3=0\Leftrightarrow x=\frac{3}{2}\)
b) \(\left(5-2x\right)\left(2x+7\right)=4x^2-25\)
\(\Leftrightarrow\left(5-2x\right)\left(2x+7\right)+\left(25-4x^2\right)=0\)
\(\Leftrightarrow\left(5-2x\right)\left(2x+7\right)+\left(5-2x\right)\left(5+2x\right)=0\)
\(\Leftrightarrow\left(5-2x\right)\left(2x+7+5+2x\right)=0\)
\(\Leftrightarrow\left(5-2x\right)\left(4x+12\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=\frac{5}{2}\\x=-3\end{array}\right.\)
c)\(x^3+27+\left(x+3\right)\left(x-9\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^2-3x+9\right)+\left(x+3\right)\left(x-9\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^2-3x+9+x-9\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^2-2x\right)=0\)
\(\Leftrightarrow\left(x+3\right)x\left(x-2\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=-3\\x=0\\x=2\end{array}\right.\)
d) \(4\left(2x+7\right)^2-9\left(x+3\right)^2=0\)
\(\Leftrightarrow\left[2\left(2x+7\right)-3\left(x+3\right)\right]\left[2\left(2x+7\right)+3\left(x+3\right)\right]=0\)
\(\Leftrightarrow\left(4x+14-3x-9\right)\left(4x+14+3x+9\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(7x+23\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=-5\\x=-\frac{23}{17}\end{array}\right.\)
đạng phân tich đa thức thành nhân tử nha các bn