![](https://rs.olm.vn/images/avt/0.png?1311)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) (x-2)3+6(x+1)2-x3+12=0
\(\Rightarrow\)x3-6x2+12x-8+6(x2+2x+1)-x3+12=0
\(\Rightarrow\)x3-6x2+12x-8+6x2+12x+6-x3+12=0
\(\Rightarrow\)24x+10=0
\(\Rightarrow\)24x=-10
\(\Rightarrow\)x=\(\dfrac{-10}{24}=\dfrac{-5}{12}\)
b)(x-5)(x+5)-(x+3)2+3(x-2)2=(x+1)2-(x-4)(x+4)+3x2
\(\Rightarrow\)x2-25-(x2+6x+9)+3(x2-4x+4)=x2+2x+1-(x2-16)+3x2
\(\Rightarrow\)x2-25-x2-6x-9+3x2-12x+12=x2+2x+1-x2+16+3x2
\(\Rightarrow\)3x2-18x-22=3x2+2x+17
\(\Rightarrow\)3x2-18x-22-3x2-2x-17=0
\(\Rightarrow\)-20x-39=0
\(\Rightarrow\)-20x=39
\(\Rightarrow\)x=\(-\dfrac{39}{20}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Về học lại hằng đẳng thức nha .-.
\(\Leftrightarrow\dfrac{\left(x+2\right)+5}{2-x}=\dfrac{2x-3}{\left(x-2\right)\left(x+2\right)}\\ \Leftrightarrow-\left(x+2\right)+5\left(x+2\right)=2x-3\\ \Leftrightarrow6x+12-2x+3=0\\ \Leftrightarrow4x+15=0\\ \Leftrightarrow x=\dfrac{-15}{4}\)
\(\dfrac{1}{x+2}+\dfrac{5}{2-x}=\dfrac{2x-3}{x^2-4}\)
\(\Leftrightarrow\dfrac{1}{x+2}-\dfrac{5}{x-2}-\dfrac{2x-3}{\left(x-2\right)\left(x+2\right)}=0\left(đk:x\ne\pm2\right)\)
\(\Leftrightarrow\dfrac{x-2-5\left(x+2\right)-2x-3}{\left(x-2\right)\left(x+2\right)}=0\)
\(\Leftrightarrow x-2-5x-10-2x-3=0\)
\(\Leftrightarrow-6x-15=0\)
\(\Leftrightarrow-6x=15\)
\(\Leftrightarrow x=-\dfrac{15}{6}\left(n\right)\)
Vậy \(S=\left\{-\dfrac{15}{6}\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{1}{x+2}+\dfrac{5}{2-x}=\dfrac{2x-3}{x^2-4}\) đkxđ : x khác 2 , x khác -2.
<=> \(\dfrac{1}{x+2}-\dfrac{5}{x-2}-\dfrac{2x-3}{x^2-4}=0\)
<=> \(\dfrac{1.\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}-\dfrac{5.\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{2x-3}{\left(x-2\right)\left(x+2\right)}=0\)
<=> \(\dfrac{x-2}{\left(x+2\right)\left(x-2\right)}-\dfrac{5x+10}{\left(x-2\right)\left(x+2\right)}-\dfrac{2x-3}{\left(x-2\right)\left(x+2\right)}=0\)
<=>\(x-2-5x-10-2x+3=0\)
<=> \(-6x-9=0\)
<=> \(x=-\dfrac{9}{6}=-\dfrac{3}{2}\left(nhận\right)\)
Vậy pt có nghiệm \(S=\left\{-\dfrac{3}{2}\right\}\)
Bài 1: Tìm x
a) (x-5) (x-3)+ 2(x-5)=0
b) (x-2)(x^2+2x+4)-(x+2)(x^2-2x+4)=2(x+2)
giúp e với ạ, e cảm ơn
![](https://rs.olm.vn/images/avt/0.png?1311)
a) (x - 5)(x - 3) + 2(x - 5) = 0
(x - 5)(x - 3 + 2) = 0
(x - 5)(x - 1) = 0
x - 5 = 0 hoặc x - 1 = 0
*) x - 5 = 0
x = 5
*) x - 1 = 0
x = 1
Vậy x = 1; x = 5
b) (x - 2)(x² + 2x + 4) - (x + 2)(x² - 2x + 4) = 2(x + 2)
x³ - 8 - x³ - 8 = 2x + 4
2x = -8 - 8 - 4
2x = -20
x = -20 : 2
x = -10
a)
\(\left(x-5\right)\left(x-3\right)+2\left(x-5\right)=0\)
\(\left(x-5\right)\left(x-3+2\right)=0\)
\(\left(x-5\right)\left(x-1\right)=0\)
\(x-5=0\) hoặc \(x-1=0\)
+) \(x-5=0\\ \Rightarrow x=5\)
+) \(x-1=0\\ \Rightarrow x=1\)
Vậy \(x=1\) hoặc \(x=5\)
b) \(\left(x-2\right)\left(x^2+2x+4\right)-\left(x+2\right)\left(x^2-2x+4\right)=2\left(x+2\right)\)
\(x^3-8-x^3-8=2x+4\)
\(2x=-8-8-4\)
\(2x=-20\)
\(x=-20:2\)
\(x=-10\)
Vậy \(x=-10\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x=\dfrac{1}{y}\Rightarrow\dfrac{1}{y}-y=4\\ \Rightarrow y^2+4y-1=0\\ \Leftrightarrow\left[{}\begin{matrix}y=-2-\sqrt{5}\Rightarrow x=2-\sqrt{5}\\y=-2+\sqrt{5}\Rightarrow x=2+\sqrt{5}\end{matrix}\right.\)
Với \(x=2-\sqrt{5};y=-2-\sqrt{5}\)
\(A=x^2+y^2=18\\ B=x^3-y^3=76\\ C=x^4+y^2=322\)
Với \(x=2+\sqrt{5};y=-2+\sqrt{5}\)
\(A=x^2+y^2=18\\ B=x^3-y^3=76\\ C=x^4+y^4=322\)
A=x^2+y^2
=(x-y)^2+2xy
=4^2+2=18
B=(x-y)^3+3xy(x-y)
=4^3+3*1*4
=64+12=76
C=(x^2+y^2)^2-2x^2y^2
=18^2-2
=322
![](https://rs.olm.vn/images/avt/0.png?1311)
a, (x-1).(x-2).(x-3)
= (x2 - 2x - x + 2) . (x-3)
= (x2 - 3x + 2). (x-3)4
= x3 - 3x2 - 3x2 + 9x + 2x -6
= x3 - 6x2 + 11x -6
b) (x2 +x+1)(x2-1)(x2-x+1)
= (x4 - x2 + x3 - x+ x2 -1) . (x2 - x +1)
= (x4 + x3 -x -1) . (x2 - x +1)
= x6 - x5 + x4 + x5 - x4 + x3 - x2 + x -1
= x6 + x3 - x2 + x - 1
c) (2x-5)(4-3x)-(3x+11)(5-2x)-15(2x-5)
= (8x - 6x2 - 20 + 15x) - (15x-6x+55-22x) - 30x + 75
= 8x - 6x2 - 20 + 15x - 15x+6x-55+22x - 30x+75
= 6x-6x2 +55
d)(x2-2x+3)(3x-5)-(x2+x-1)(2x+7)
làm tương tự phần C
lưu ý trước dấu ngoặc là dấu trừ, khi phá ngoặc ra phải đổi dấu
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\frac{-x^2y^5}{-x^2y^5}=1\)
b)\(\frac{-\left(x^7y^5z\right)^2}{-\left(xy^3z\right)^2}=\frac{x^{14}y^{10}z^2}{x^2y^6z^2}=x^7.y^4\)Thế vào ta được 1.(-10)^4=10000 cái khi nãy làm lộn
câu a cả tử và mẫu đều giống nhau nên kết quả là 1
b) chia ra ta được x6y2. Thế vào thì ra 1.102=100
E=x5-5x4+5x3-5x2+5x-1
=x5-4x4+x3-4x2+x-x4+4x3-x2+4x-1
=x(x4-4x3+x2-4x+1)-(x4-4x3+x2-4x+1)
=(x-1)(x4-4x3+x2-4x+1)
Tại x=4 ta có:
E=(4-1)(44-4*43+42-4*4+1)
=3*(256-256+16-16+1)
=3*1
=3
Thanks! Kết bn đcj ko dzậy![vui vui](/media/olmeditor/plugins/smiley/images/vui.png)