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câu a tìm mẫu thức chung, rồi đk là mtc khác 0
câu b rút dễ mà
câu c đập vô thôi
Bài 4:
\(b,\dfrac{\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}}{1+\dfrac{x^3}{1-x^3}}\)
\(=\dfrac{\dfrac{\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}-\dfrac{\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}}{\dfrac{1-x^3}{1-x^3}+\dfrac{x^3}{1-x^3}}\)
\(=\dfrac{\dfrac{x^2+2x+1-x^2+2x-1}{\left(x-1\right)\left(x+1\right)}}{\dfrac{1-x^3+x^3}{1-x^3}}\)
\(=\dfrac{4x}{\left(x-1\right)\left(x+1\right)}:\dfrac{1}{\left(1-x\right)\left(1+x+x^2\right)}\)
\(=\dfrac{4x}{\left(x-1\right)\left(x+1\right)}:\dfrac{-1}{\left(x-1\right)\left(1+x+x^2\right)}\)
\(=\dfrac{4x}{\left(x-1\right)\left(x+1\right)}.\left[-\left(x-1\right)\left(x^2+x+1\right)\right]\)
\(=\dfrac{-4x\left(x^2+x+1\right)}{x+1}\)
Bài 4:
=>x(x^2+1)=0
=>x=0
Bài 5:
=>\(3n^3+n^2+9n^2-1-4⋮3n+1\)
=>\(3n+1\in\left\{1;-1;2;-2;4;-4\right\}\)
hay \(n\in\left\{0;-1;1\right\}\)
Bài 3:
\(a^3+b^3+c^3-3bac\)
\(=\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc\)
\(=\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2\right)-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)=0\)
Bài 2:
a: \(M=\dfrac{x^2+2x}{2\left(x+5\right)}+\dfrac{x-5}{x}+\dfrac{50-5x}{2x\left(x+5\right)}\)
\(=\dfrac{x^3+2x^2+2\left(x-5\right)\left(x+5\right)+50-5x}{2x\left(x+5\right)}\)
\(=\dfrac{x^3+2x^2+50-5x+2x^2-50}{2x\left(x+5\right)}\)
\(=\dfrac{x^3+4x^2-5x}{2x\left(x+5\right)}=\dfrac{x\left(x^2+4x-5\right)}{2x\left(x+5\right)}=\dfrac{x-1}{2}\)
b: Khi x=3 thì \(M=\dfrac{3-1}{2}=\dfrac{2}{2}=1\)
Khi x=5 thì \(M=\dfrac{5-1}{2}=\dfrac{4}{2}=2\)
a) 2x + 1 = 5 - 5x
=> 2x + 5x = 5 - 1
=> 7x = 4
=> x = 4/7
b) 3x - 2 = 2x + 5
=> 3x - 2x = 5 + 2
=> x = 7
c) 7(x - 2) = 5(3x + 1)
=> 7x - 14 = 15x + 5
=> 7x - 15x = 5 + 14
=> - 8x = 19
=> x = - 19/8
d) 2x + 5 = 20 - 3x
=> 2x + 3x = 20 - 5
=> 5x = 15
=> x = 3
e) x - 3 = 18 - 5x
=> x + 5x = 18 + 3
=> 6x = 21
=> x = 21/6 = 7/2
Dạng 1: Rút gọn
Bài 1
a) Rút gọn
P= (\(\dfrac{8}{x^2-16}+\dfrac{1}{x+4}\)):\(\dfrac{1}{x^2-2x-8}\)
= (\(\dfrac{8}{\left(x+4\right)\left(x-4\right)}+\dfrac{x-4}{\left(x+4\right)\left(x-4\right)}\)):\(\dfrac{1}{\left(x-4\right)\left(x+2\right)}\)
= \(\dfrac{x+4}{\left(x+4\right)\left(x-4\right)}:\dfrac{1}{\left(x-4\right)\left(x+2\right)}\)
= \(\dfrac{1}{x-4}.\left(x-4\right)\left(x+2\right)\)
= x+2
Bài 2
a) Rút gọn
D=(\(\dfrac{1}{x-1}-\dfrac{x}{1-x^3}.\dfrac{x^2+x+1}{x+1}\)):\(\dfrac{2x+1}{x^2+x+1}\)
= (\(\dfrac{1}{x-1}+\dfrac{x}{\left(x-1\right)\left(x+1\right)}\)):\(\dfrac{2x+1}{x^2+x+1}\)
= \(\dfrac{2x+1}{\left(x+1\right)\left(x-1\right)}\).\(\dfrac{x^2+x+1}{2x+1}\)
= \(\dfrac{x^2+x+1}{\left(x+1\right)\left(x-1\right)}\)
b) Tìm x∈Z để D∈Z
D=\(\dfrac{x^2+x+1}{\left(x+1\right)\left(x-1\right)}=\dfrac{x^2+x+1}{x^2-1}=\dfrac{x^2-1}{x^2-1}+\dfrac{x+2}{x^2-1}=1+\dfrac{x+2}{x^2-1}\)Để D nguyên thì x+2=0⇔x=-2(t/m)
Vậy ...........................
Dạng 2: Phương trình
Bài 1. Giải phương trình
a) 2x+1=5-5x
⇔ 2x+5x=5-1
⇔ 7x=4
⇔ x=\(\dfrac{4}{7}\)
Vậy S=\(\left\{\dfrac{4}{7}\right\}\) là tập nghiệm của hương trình
b) 3x-2=2x+5
⇔ 3x-2x=5+2
⇔ x=7
Vậy......................
c) 7(x-2)=5(3x+1)
⇔ 7x-14=15x+5
⇔ 7x-15x=5+14
⇔ -8x=19
⇔ x=-\(\dfrac{19}{8}\)
Vậy..........................
d) 2x+5=20-3x
⇔ 2x+3x=20-5
⇔ 5x=15
⇔ x=3
Vậy...................
e) x-3=18-5x
⇔ x+5x=18+3
⇔ 6x=21
⇔ x=\(\dfrac{7}{2}\)
Vậy..............................
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)
\((3x-5)(2x+11) -(2x+3)(3x+7) \)
\(= 6x^2 +33x-10x-55-6x^2-14x-9x-10\)
\(= (6x^2-6x^2)+(33x-10x-14x-9x)-(55+10)\)
\(=-65\)
\(\)
a) \(A=\left(x-2\right)^2-\left(x-5\right)\left(x+5\right)\)
\(=\left(x^2-4x+4\right)-\left(x^2-25\right)\)
\(=29-4x\)
b) \(A=29-4x=1\)
\(\Leftrightarrow4x=28\)
\(\Leftrightarrow x=7\)