câu này 4(x 2) x^2 2x=0
là (x-2) hay (x+2) a
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Bài 6 : Tìm x, biết :
a, ( 2x-5)^2 - ( 5+2x)^2 = 0
b, 27x^3 - 54x^2 + 36x = 8
c, ( x^3 + 8 ) - ( x+2) ( x-4) = 0
d, x^6 - 1 = 0
Giúp mk vs ạ mk đang cần gấp
Tìm x,y nguyên:
1) (2x-1).(2x-5)<0
2) (3x+1).(5-2x)>0
3) (3-2x).(x+2)>0
4) (2-x).(x+1)=|y+1|
5) (x+3).(1-x)=|y|
6) (x-2).(5-x)=|2y+1|+2
7)(x-3).(x-5)+|y-2|=0
8) (x-2).(5-x)-|y+1|=1
GIÚP MK VS MK TICK CHO
THANK FOR WATCHING!
giải các phương trình tích sau:
1, 3x(x-2) = 7(x-2)
2, 2x^2 = x
3, x^2(x^2+1) = 0
4, x^3+9x = 6x^2
5, (x+3)(x-3) = 16
6, (x-6)(x+4) = 2(x+1)
7, (x-1)^2 = 4
8, (2x+1)^2 = (x-1)^2
9,(x^2-1)(2x-1) = (x^2-1)(x+2)
10, x^2-9x+20 = 0
11, x^2+2x-15 = 0
12, x^3-4x^2+5x = 0
13,x^3+4x^2+x-6 = 0
14, x^3-3x^2+4 = 0
15, x^4+2x^3+2x^2-2x-3 = 0
16, (x^2+x)(x^2+x+1) = 6
mk cần gấp mai mk đi học
1)3x(x-2)=7(x-2)
<=>3x(x-2)-7(x-2)=0
<=>(x-2)(3x-7)=0
x-2=0=>x=2
3x-7=0=>x=7/3
cn lại lm tg tự
10)\(x^2-9x+20=0\)
\(\Leftrightarrow x^2-4x-5x+20=0\)
\(\Leftrightarrow x\left(x-4\right)-5\left(x-4\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(x-5\right)=0\)
\(\Leftrightarrow\hept{\begin{cases}x=4\\x=5\end{cases}}\)
16) \(\left(x^2+x\right)\left(x^2+x+1\right)=6\)
\(\Leftrightarrow x^4+x^3+x^2+x^3+x^2+x=6\)
\(\Leftrightarrow x^4+2x^3+2x^2+x-6=0\)
\(\Leftrightarrow x^4+2x^3+2x^2+4x-3x-6=0\)
\(\Leftrightarrow x^3\left(x+2\right)+2x\left(x+2\right)-3\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x^3+2x-3\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x^3+\frac{1}{4}x-x+\frac{11}{4}x-\frac{11}{4}-\frac{1}{4}+x^2-x^2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left[\left(x^3-x^2\right)+\left(x^2-x\right)+\left(\frac{1}{4}x-\frac{1}{4}\right)+\left(\frac{11}{4}x-\frac{11}{4}\right)\right]=0\)
\(\Leftrightarrow\left(x+2\right)\left[x^2\left(x-1\right)+x\left(x-1\right)+\frac{1}{4}\left(x-1\right)+\frac{11}{4}\left(x-1\right)\right]=0\)
\(\Leftrightarrow\left(x+2\right)\left(x-1\right)\left(x^2+x+\frac{1}{4}+\frac{11}{4}\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x-1\right)\left[\left(x+\frac{1}{2}\right)^2+\frac{11}{4}\right]=0\)
\(\Leftrightarrow\hept{\begin{cases}x+2=0\\x-1=0\\\left(x+\frac{1}{2}\right)^2+\frac{11}{4}=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x=-2\\x=1\\\left(x+\frac{1}{2}\right)^2+\frac{11}{4}=0->ktm\end{cases}}\)
\(\left(x+\frac{1}{2}\right)^2+\frac{11}{4}\ge\frac{11}{4}>0\)=>ko thỏa mãn(đây là giải thích cho phần trên)
6)\(\left(x-6\right)\left(x+4\right)=2\left(x+1\right)\)
\(\Leftrightarrow x^2+4x-6x-24-2x-2=0\)
\(\Leftrightarrow x^2-4x-26=0\)
đến đây nếu phân tích tam thức bậc hai này thì tìm đc x là số thập phân vô hạn ko tuần hoàn nên mk nghĩ là đề bài câu này sai
Câu 1. Giải các phườn trình sau:
a, 3x+6=0
b, 2x-10=0
c, 3x-7=11
d, 3x-9=0
e, 3x(2-x) =15(x-2)
f, (x+5)(x+4)=0
g, x(x+4)=0
h, (2x -4)(x-2)=0
i, (x+1/5)(2x-3)=0
k, x²-4x=0
m, 4x²-1=0
n, x²-6x+9=0
l, (3x-5)²-(x+4)²=0
o, 7x(x+2)-5(x+2)=0
p, 3x(2x-5)-4x+10=0
q, (2-2x)-x²+1=0
r, x(1-3x)=5(1-3x)
s, 2x-3/4+x+1/6=3
t, x-3/4-2x+1/3=x/6
u, x+1/13+x+2/12=x+3/11+x+4/10
v, 2x+1/15+2x+2/14=2x+3/13+2x+4/12
Giúp e nha mn. E cảm ơn trc ạ!
e, 3x(2-x) =15(x-2)
\(\Leftrightarrow3x\left(2-x\right)-15\left(x-2\right)=0\)
\(\Leftrightarrow-3x\left(x-2\right)-15\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(-3x-15\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-2=0\\-3x-15=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\x=5\end{matrix}\right.\)
Vậy..
f, (x+5)(x+4)=0
\(\Leftrightarrow\left\{{}\begin{matrix}x+5=0\\x+4=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-5\\x=-4\end{matrix}\right.\)
Vậy..
g, x(x+4)=0
\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x+4=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=-4\end{matrix}\right.\)
,h, (2x -4)(x-2)=0
\(\Leftrightarrow2\left(x-2\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(2-1\right)=0\)
\(\Leftrightarrow x-2=0\Leftrightarrow x=2\)
i, (x+1/5)(2x-3)=0
\(\Leftrightarrow\left\{{}\begin{matrix}x+\frac{1}{5}=0\\2x-3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\frac{-1}{5}\\x=\frac{3}{2}\end{matrix}\right.\)
k, x²-4x=0
\(\Leftrightarrow x\left(x-2\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)
m, 4x²-1=0
\(\Leftrightarrow\left(2x\right)^2-1^2=0\)
\(\Leftrightarrow\left(2x-1\right)\left(2x+1\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x-1=0\\2x+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=1\\2x=-1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{1}{2}\\x=\frac{-1}{2}\end{matrix}\right.\)
n, x²-6x+9=0
\(\Leftrightarrow x^2-2.x.3+3^2=0\)
\(\Leftrightarrow\left(x-3\right)^2=0\Leftrightarrow x-3=0\)
<=> x=3
l, (3x-5)²-(x+4)²=0
\(\Leftrightarrow\left(3x-5-x-4\right)\left(3x-5+x+4\right)=0\)
\(\Leftrightarrow\left(2x-9\right)\left(4x-1\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x-9=0\\4x-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=9\\4x=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{9}{2}\\x=\frac{1}{4}\end{matrix}\right.\)
Vậy ..
o, 7x(x+2)-5(x+2)=0
\(\Leftrightarrow\left(x+2\right)\left(7x-5\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+2=0\\7x-5=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-2\\7x=5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-2\\x=\frac{5}{7}\end{matrix}\right.\)
Vậy....
p, 3x(2x-5)-4x+10=0
\(\Leftrightarrow3x\left(2x-5\right)-\left(4x-10\right)=0\)
\(\Leftrightarrow3x\left(2x-5\right)-2\left(2x-5\right)=0\)
\(\Leftrightarrow\left(2x-5\right)\left(3x-2\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x-5=0\\3x-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=5\\3x=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{5}{2}\\x=\frac{2}{3}\end{matrix}\right.\)
Vậy...
q, (2-2x)-x²+1=0
\(\Leftrightarrow2\left(1-x\right)-\left(x^2-1^2\right)=0\)
\(\Leftrightarrow2\left(1-x\right)-\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow2\left(1-x\right)+\left(1-x\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left(1-x\right)\left(2+x+1\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}1-x=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-1\\x=-3\end{matrix}\right.\)
Vậy ....
r, x(1-3x)=5(1-3x)
\(\Leftrightarrow x\left(1-3x\right)-5\left(1-3x\right)=0\)
\(\Leftrightarrow\left(1-3x\right)\left(x-5\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}1-3x=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-3x=-1\\x=5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{1}{3}\\x=5\end{matrix}\right.\)
s, 2x-3/4+x+1/6=3
\(\Leftrightarrow x-\frac{7}{12}=3\Leftrightarrow x=3+\frac{7}{12}=\frac{43}{12}\)
r, x(1-3x)=5(1-3x)
➜x(1-3x)-5(1-3x)=0
➜(x-5)(1-3x)=0
➜\(\left[{}\begin{matrix}x-5=0\\1-3x=0\end{matrix}\right.\text{➜}\left[{}\begin{matrix}x=5\\x=\frac{1}{3}\end{matrix}\right.\)
Mk lười lắm mai nha!!!~~~~~~~~~~~~
Làm dần:
a, 3x+6=0
➜3x=-6
➜x=2
b, 2x-10=0
➜2x=10
➜x=5
c, 3x-7=11
➜3x=11+7
➜3x=18
➜x=6
d, 3x-9=0
➜3x=9
➜x=3
I. Tìm x biết
a) ( x - 3 )2 - 4 = 0
b) x2 - 2x = 24
c) ( 2x + 1 )2 + ( x + 3 )2 - 5 ( x - 7 ) ( x + 7 ) = 0
d) ( x - 3 ) ( x2 + 3x + 9 ) + x ( x + 2 ) ( 2 - x ) = 1
e) ( 3x - 1 )2 + 2 ( x + 3 )2 + 11 ( x + 1 ) ( 1 - x ) = 6
Các bạn giải chi tiết giúp mk nha mk đang cần gấp!!!!
\(a,\left(x-3\right)^2-4=0\)
\(\Leftrightarrow\left(x-3\right)^2=4\)
\(\Rightarrow x-3=\pm2\)
\(\hept{\begin{cases}x-3=2\Rightarrow x=5\\x-3=-2\Rightarrow x=1\end{cases}}\)
Vậy \(x=5\)hoặc \(x=1\)
\(b,x^2-2x=24\)
\(\Leftrightarrow x^2-2x+1-1=24\)
\(\Leftrightarrow\left(x-1\right)^2=24+1=25\)
\(\Leftrightarrow x-1=\pm5\)
\(\hept{\begin{cases}x-1=5\Rightarrow x=6\\x-1=-5\Rightarrow x=-4\end{cases}}\)
Vậy \(x=6\) hoặc \(x=-4\)
\(c,\left(2x+1\right)^2+\left(x+3\right)^2-5\left(x-7\right)\left(x+7\right)=0\)
\(\Leftrightarrow4x^2+4x+1+x^2+6x+9-5\left(x^2-49\right)=0\)
\(\Leftrightarrow4x^2+4x+1+x^2+6x+9-5x^2+245=0\)
\(\Leftrightarrow10x+255=0\)
\(\Leftrightarrow10x=-255\)
\(\Leftrightarrow x=\frac{-51}{2}\)
\(d,\left(x-3\right)\left(x^2+3x+9\right)+x\left(x+2\right)\left(2-x\right)=1\)
\(\Leftrightarrow x^3-27+x\left(2x-x^2+4-2x\right)=1\)
\(\Leftrightarrow x^3-27-x^3+4x=1\)
\(\Leftrightarrow4x-27=1\)
\(\Leftrightarrow4x=28\)
\(\Leftrightarrow x=7\)
\(e,\left(3x-1\right)^2+2\left(x+3\right)^2+11\left(x+1\right)\left(1-x\right)=6\)
\(\Leftrightarrow9x^2-6x+1+2\left(x^2+6x+9\right)+11\left(x-x^2+x-x\right)=6\)
\(\Leftrightarrow9x^2-6x+1+2x^2+12x+18-11x^2+11x=6\)
\(\Leftrightarrow17x+19=6\)
\(\Leftrightarrow17x=-13\)
\(\Leftrightarrow x=\frac{-13}{17}\)
Bài 1: tìm x b) x-1,4/1,4-x =2 c) /x-1/ + (2-3x) < 0 d) /-1/2x + 3 / <= 5
Bài 2: tìm x a) /x-1/ + (2-3x) < 0
b) /x-1/ = (-x) - 5
c) /x-2/ + /x-4/ =5
d) /x-1/ + /x-3/ =6
e) / x-1/ + (x+3)
*Các bạn giúp mk với, Thanks nhiều ạ. Các bạn làm đầy đủ thì mình cảm ơn lắm!!!
Bài 1: tìm x b) x-1,4/1,4-x =2 c) /x-1/ + (2-3x) < 0 d) /-1/2x + 3 / <= 5
Bài 2: tìm x a) /x-1/ + (2-3x) < 0
b) /x-1/ = (-x) - 5
c) /x-2/ + /x-4/ =5
d) /x-1/ + /x-3/ =6
e) / x-1/ + (x+3)
*Các bạn giúp mk với, Thanks nhiều ạ. Các bạn làm đầy đủ thì mình cảm ơn lắm!!!
Các bạn giải rõ ràng ra giúp mình nhé, làm bao nhiêu câu cx đc, hết càng tốt. Ngày mai mình nộp bài gòi. Thx các bạn trc <3
Bài: Tìm x, biết:
1/ (x - 6).(2x + 10) > 0
2/ (2x - 8).(x - 2) < 0
3/ (3x - 1) / (x + 2) < hoặc = 0
4/ (4x + 3) / (2x - 5) > 0
5/ (x + 5) / (x + 3) < 1
6/ (x - 7)^x+1 = (x - 7)^x+11