K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
HA
Tìm x biết:
a, ( 2x - 5)2 = ( x - 2)2
b, ( x + 1)( 2 - x) - ( 3x + 5)( x + 2) = - 4x2 + 1
Help me!!!
4
22 tháng 10 2017
\(a,\left(2x-5\right)^2=\left(x-2\right)^2\)
\(\Rightarrow\left(2x-5\right)^2-\left(x-2\right)^2=0\)
\(\Rightarrow\left(2x-5-x+2\right)\left(2x-5+x-2\right)=0\)
\(\Rightarrow\left(x-3\right)\left(3x-7\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-3=0\\3x-7=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{7}{3}\end{matrix}\right.\)
\(b,\left(x+1\right)\left(2-x\right)-\left(3x+5\right)\left(x+2\right)=-4x^2+1\)
\(\Rightarrow2x-x^2+2-x-3x^2-6x-5x-10=-4x^2+1\)
\(\Rightarrow-10x-4x^2-12=-4x^2+1\)
\(\Rightarrow-10x-4x^2-12+4x^2-1=0\)
\(\Rightarrow-10x-13=0\)
\(\Rightarrow x=-\dfrac{13}{10}\)
15 tháng 2 2020
\(\frac{3x+2}{x-1}+\frac{2x-4}{x+2}=5\)
\(\Rightarrow\frac{\left(3x+2\right)\left(x+2\right)+\left(2x-4\right)\left(x-1\right)}{\left(x-1\right)\left(x+2\right)}=5\)
\(\Rightarrow\frac{3x^2+2x+6x+4+2x^2-4x-2x+4}{\left(x-1\right)\left(x+2\right)}=5\)
\(\Rightarrow\frac{5x^2+2x+8}{\left(x-1\right)\left(x+2\right)}=5\)
\(\Rightarrow5x^2+2x+8=5\left(x-1\right)\left(x+2\right)\)
\(\Rightarrow5x^2+2x+8=5x^2-5x+10x-10\)
\(\Rightarrow5x^2-5x^2+2x-5x=8-10\)
\(\Rightarrow-3x=-2\)
\(\Rightarrow x=\frac{2}{3}\)
1 tháng 4 2020
\( a)2x - 3 = 3x - 7\\ \Leftrightarrow 2x - 3x = - 7 + 3\\ \Leftrightarrow - x = - 4\\ \Leftrightarrow x = 4\\ b)x - \left( {6 - 5x} \right) = 2\left( {x - 1} \right) + 12\\ \Leftrightarrow 6x - 6 = 2x - 2 + 12\\ \Leftrightarrow 6x - 2x = 10 + 6\\ \Leftrightarrow 4x = 16\\ \Leftrightarrow x = 4\\ c){x^4} - 144x = 2{x^2} + 1295\\ \Leftrightarrow {x^4} - 2{x^2} - 144x - 1295 = 0\\ \Leftrightarrow \left( {x + 5} \right)\left( {{x^3} - 5{x^2} + 23x - 259} \right) = 0\\ \Leftrightarrow \left( {x + 5} \right)\left( {x - 7} \right)\left( {{x^2} + 2x + 37} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l} x = - 5\\ x = 7\\ {x^2} + 2x + 37 = 0\left( {vn} \right) \end{array} \right. \)
LT
1 tháng 4 2020
a) \(2x-3=3x-7\)
\(\Leftrightarrow x=4\)
b) \(x-\left(6-5x\right)=2\left(x-1\right)+12\)
\(\Leftrightarrow x-6+5x=2x-2+12\)
\(\Leftrightarrow\)\(4x=16\)
\(\Leftrightarrow x=4\)
c) \(x^4-144x=x^2+1295\)
\(\Leftrightarrow x^4-x^2-144x-1295=0\)
\(\Leftrightarrow\left(x^4+2x^2+1\right)-\left(4x^2+144x+1295\right)=0\)
\(\Leftrightarrow\left(x^2+1\right)^2-\left(2x+36\right)^2=0\)
\(\Leftrightarrow\left(x^2+1+2x+36\right)\left(x^2+1-2x-36\right)=0\)
\(\Leftrightarrow\left(x^2+2x+37\right)\left(x^2-2x-35\right)=0\)
\(\Leftrightarrow\left(x^2+2x+1+36\right)\left(x^2+2x-7x-35\right)=0\)
\(\Leftrightarrow\left[\left(x+1\right)^2+36\right]\left[\left(x+5\right)\left(x-7\right)\right]=0\)
do \(\left(x+1\right)^2+36\ge36\forall x\)
\(\Rightarrow\left[{}\begin{matrix}x+5=0\\x-7=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-5\\x=7\end{matrix}\right.\)
Vậy/...
D
20 tháng 11 2019
⇔ \((\frac{3x}{x+3}+\frac{2}{x-5}):\frac{1}{\left(x-5\right)\left(x+3\right)}\)
ĐK : x \(\ne-3,\) x \(\ne5\)
\(\Leftrightarrow\left[\frac{3x\left(x-5\right)}{\left(x+3\right)\left(x-5\right)}+\frac{2\left(x+3\right)}{\left(x-5\right)\left(x+3\right)}\right]:\frac{1}{\left(x+3\right)\left(x-5\right)}\)
\(\Leftrightarrow\left[\frac{3x^2-15x+2x+6}{\left(\right)\left(\right)}\right]:\frac{1}{\left(\right)\left(\right)}\)
\(\Leftrightarrow\left[\frac{3x^2-13x+6}{\left(x-5\right)\left(x+3\right)}\right].\left(x+3\right)\left(x-5\right)\)
\(\Leftrightarrow3x^2-13x+6\)
DH
27 tháng 7 2016
bài 1:
a. \((x+1)(x+3) - x(x+2)=7 \)
\(x^2+ 3x +x +3 - x^2 -2x =7\)
\(x^2+4x+3-x^2-2x=7\)
\(=> 2x+3=7\)
\(2x=4\)
\(x = 2\)
Bài 2:
a)
\((3x-5)(2x+11) -(2x+3)(3x+7) \)
\(= 6x^2 +33x-10x-55-6x^2-14x-9x-10\)
\(= (6x^2-6x^2)+(33x-10x-14x-9x)-(55+10)\)
\(=-65\)
\(\)
15 tháng 8 2017
Bài 2 :
Câu a : \(y\left(y^3+y^2-y-2\right)-\left(y^2-2\right)\left(y^2+y+1\right)\)
\(=y^4+y^3-y^2-2y-y^4-y^3-y^2+2y^2+2y+2\)
\(=2\) \(\Rightarrow\) ko phụ thuộc vào biến .
Câu b : \(\left(2x+3\right)\left(4x^2-6x+9\right)-2\left(4x^3-1\right)\)
\(=8x^3-12x^2+18x+12x^2-18x+27-8x^3+2\)
\(=29\Rightarrow\) ko thuộc vào biến
Câu c : \(3x\left(x+5\right)-\left(3x+18\right)\left(x-1\right)\)
\(=3x^2+15x-3x^2+3x-18x+18\)
\(=18\) \(\Rightarrow\) ko thuộc vào biến
Câu d : \(\left(2x+6\right)\left(4x^2-12x+36\right)-8x^3+5\)
\(=8x^3-24x^2+72x+24x^2-72x+216-8x^3+5\)
\(=221\) \(\Rightarrow\) không thuộc vào biến
16 tháng 8 2017
câu 1) a) \(\left(x^2+2xy+y^2\right)\left(x+y\right)=\left(x+y\right)^2\left(x+y\right)=\left(x+y\right)^3\)
b) \(y\left(y^3+y^2-3y-2\right)+\left(y^2-2\right)\left(y^2+y-1\right)\)
\(=y^4+y^3-3y^2-2y+y^4+y^3-y^2-2y^2-2y+2\)
\(=2y^4+2y^3-6y^2-4y+2=2y\left(y^3+y^2-3y-2\right)+2\)
\(=2y\left(y+2\right)\left(y^2-y-1\right)+2=2\left(y^2+2y\right)\left(y^2-y-1\right)+2\)
\(=2\left(y^2+2y\right)\left(y^2-y-1+1\right)=2\left(y^2+2y\right)\left(y^2-y\right)\)
c) \(6x^2-\left(2x+5\right)\left(3x-2\right)=6x^2-\left(6x^2-4x+15x-10\right)\)
\(\Leftrightarrow6x^2-6x^2+4x-15x+10=-11x+10\)
d) \(\left(2x-1\right)\left(3x+1\right)+\left(3x+4\right)\left(3-2x\right)\)
\(\)\(=6x^2+2x-3x-1+9x-6x^2+12-8x=11\)
e) \(\left(3x-5\right)\left(7-5x\right)-\left(5x+2\right)\left(2-3x\right)\)
\(=21x-15x^2-35+25x-\left(10x-15x^2+4-6x\right)\)
\(21x-15x^2-35+25x-10x+15x^2-4+6x=42x-39\)