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H
4 tháng 8 2019
\(\frac{3}{x+1}+\frac{2}{x+2}=\frac{5x+4}{x^2+3x+2}.\)ĐKXĐ: \(x\ne-1;-2\)
\(\Leftrightarrow\frac{3\left(x+2\right)}{\left(x+1\right)\left(x+2\right)}+\frac{2\left(x+1\right)}{\left(x+1\right)\left(x+2\right)}=\frac{5x+4}{\left(x+1\right)\left(x+2\right)}\)
\(\Leftrightarrow3x+6+2x+2=5x+4\)
\(\Leftrightarrow3x+2x-5x=-6-2+4\)
\(\Leftrightarrow0x=-4\)
=> PT vô nghiệm
\(2;\frac{2}{3x-1}-\frac{15}{6x^2-x-1}=\frac{3}{2x-1}\)
\(\Leftrightarrow\frac{2\left(2x-1\right)}{\left(2x-1\right)\left(3x-1\right)}-\frac{15}{6x^2+3x-2x-1}=\frac{3\left(3x-1\right)}{\left(2x-1\right)\left(3x-1\right)}\)
\(\Leftrightarrow\frac{4x-2-15}{\left(2x-1\right)\left(3x-1\right)}=\frac{9x-3}{\left(2x-1\right)\left(3x-1\right)}\)
\(\Leftrightarrow4x-2-15=9x-3\)
\(\Leftrightarrow4x-9x=2+15-3\)
\(\Leftrightarrow-5x=14\)
.....
DN
21 tháng 8 2018
\(A=\left(x+2\right)\left(x^2-2x+4\right)-\left(x^3-2\right)\)
\(\Rightarrow A=\left(x^3+8\right)-\left(x^3-2\right)\)
\(\Rightarrow A=x^3+8-x^3+2\)
\(\Rightarrow A=\left(x^3-x^3\right)+\left(8+2\right)\)
\(\Rightarrow A=10\)
21 tháng 8 2018
\(A=\left(x+2\right)\left(x^2-2x+4\right)-\left(x^3-2\right)\)
\(=x^3+8-x^3+2\)
\(=10\)
\(B=\left(x+2\right)\left(x-2\right)\left(x^2+2x+4\right)\left(x^2-2x+4\right)\)
\(=\left(x+2\right)\left(x^2-2x+4\right)\left(x-2\right)\left(x^2+2x+4\right)\)
\(=\left(x^3+8\right)\left(x^3-8\right)\)
\(=x^6-64\)
\(C=\left(x^2+3x+1\right)^2+\left(3x-1\right)^2-2\left(x^2+3x+1\right)\left(3x-1\right)\)
\(=\left(x^2+3x+1\right)^2-2\left(x^2+3x+1\right)\left(3x-1\right)+\left(3x-1\right)^2\)
\(=\left(x^2+3x+1-3x+1\right)^2\)
\(=\left(x^2+2\right)^2\)
\(D=\left(3x^3+3x+1\right)\left(3x^3-3x+1\right)-\left(3x^3+1\right)^2\)
\(=\left(3x^3+1+3x\right)\left(3x^3+1-3x\right)-\left(3x^3+1\right)^2\)
\(=\left(3x^3+1\right)^2-9x^2-\left(3x^3+1\right)^2\)
\(=-9x^2\)
\(E=\left(2x^2+2x+1\right)\left(2x^2-2x+1\right)-\left(2x^2+1\right)^2\)
\(=\left(2x^2+1+2x\right)\left(2x^2+1-2x\right)-\left(2x^2+1\right)^2\)
\(=\left(2x^2+1\right)^2-4x^2-\left(2x^2+1\right)^2\)
\(=-4x^2\)
Q
18 tháng 7 2017
a) \(\left(4x-1\right)^2-\left(3x+2\right)\left(3x-2\right)=\left(7x-1\right)\left(x+2\right)+\left(2x+1\right)^2-\left(4x^2+7\right)\)(1)
\(\Leftrightarrow\left(16x^2-8x+1\right)-\left(9x^2-4\right)=\left(7x^2+14x-x-2\right)+\left(4x^2+4x+1\right)-\left(4x^2+7\right)\)
\(\Leftrightarrow16x^2-8x+1-9x^2+4=7x^2+13x-2+4x^2+4x+1-4x^2-7\)
\(\Leftrightarrow7x^2-8x+5=7x^2+17x-8\)
\(\Leftrightarrow7x^2-8x-7x^2-17x=-8-5\)
\(\Leftrightarrow-25x=-13\)
\(\Leftrightarrow x=\dfrac{13}{25}\)
Vậy tập nghiệm phương trình (1) là \(S=\left\{\dfrac{13}{25}\right\}\)
KN
22 tháng 10 2017
Bài 1
a) (x5 + 4x3 - 6x2) : 4x2
= 4x2(\(\dfrac{1}{4}\)x3 + x - \(\dfrac{3}{2}\)) : 4x2
= \(\dfrac{1}{4}\)x3 + x - \(\dfrac{3}{2}\)
b) (x3 - 8) : (x2 + 2x + 4)
= (x - 2)(x2 + 2x + 4) : (x2 + 2x + 4)
= x - 2
c) (3x2 - 6x) : (2 - x)
= -(6x - 3x2) : (2 - x)
= -3x(2 - x) : (2 - x)
= -3x
d) (x3 + 2x2 - 2x - 1) : (x2 + 3x + 1)
= [(x3 - 1) + (2x2 - 2x)] : (x2 + 3x + 1)
= [(x - 1)(x2 + x + 1) + 2x(x - 1)] : (x2 + 3x + 1)
= (x - 1)(x2 + x + 1 + 2x) : (x2 + 3x + 1)
= (x - 1)(x2 + 3x + 1) : (x2 + 3x + 1)
= x - 1
Bài 2
a) (x - 4)2 - (x - 2)(x + 2) = 6
x2 - 8x + 16 - (x2 - 4) = 6
x2 - 8x + 16 - x2 + 4 = 6
-8x + 20 = 6
\(\Rightarrow\) -8x = - 14
\(\Rightarrow\) x = \(\dfrac{7}{4}\)
b) 9(x + 1)2 - (3x - 2)(3x + 2) = 10
9(x2 + 2x + 1) - (9x2 - 4) = 10
9x2 + 18x + 9 - 9x2 + 4 = 10
18x + 13 = 10
\(\Rightarrow\) 18x = -3
\(\Rightarrow\) x = \(\dfrac{-1}{6}\)
Nhớ tik mik nha 
không lần sau mik ko giúp đâu 
AK... có j ko hiểu thì bn cứ bình luận bên dưới 