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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..............................
Bài2:
a: \(Q=\left(\dfrac{2x+1}{x\left(x-5\right)}-\dfrac{2x}{x\left(x+5\right)}\right)\cdot\dfrac{x\left(x-5\right)\left(x+5\right)}{21x-2}\)
\(=\dfrac{2x^2+11x+5-2x^2+10x}{x\left(x-5\right)\left(x+5\right)}\cdot\dfrac{x\left(x-5\right)\left(x+5\right)}{21x-2}\)
\(=\dfrac{21x+5}{21x-2}\)
b: Để Q là số nguyên thì \(21x-2+7⋮21x-2\)
\(\Leftrightarrow21x-2\in\left\{1;-1;7;-7\right\}\)
hay \(x\in\left\{\dfrac{1}{7};\dfrac{1}{21};\dfrac{3}{7};-\dfrac{5}{21}\right\}\)
b, \(P=\dfrac{x+2}{x+3}-\dfrac{5}{x^2+3x-2x-6}+\dfrac{1}{2-x}\)
\(=\dfrac{x+2}{x+3}-\dfrac{5}{\left(x+3\right)\left(x-2\right)}-\dfrac{1}{x-2}\)
\(=\dfrac{\left(x+2\right)\left(x-2\right)}{\left(x+3\right)\left(x-2\right)}-\dfrac{5}{\left(x+3\right)\left(x-2\right)}-\dfrac{x+3}{\left(x+3\right)\left(x-2\right)}\)
\(=\dfrac{x^2-4-5-x-3}{\left(x+3\right)\left(x-2\right)}=\dfrac{x^2-x-12}{\left(x+3\right)\left(x-2\right)}\)
\(=\dfrac{x^2-4x+3x-12}{\left(x+3\right)\left(x-2\right)}\)
\(=\dfrac{\left(x-4\right)\left(x+3\right)}{\left(x+3\right)\left(x-2\right)}=\dfrac{x-4}{x-2}\)
c, Để \(P=-\dfrac{3}{4}\)
\(\Leftrightarrow\dfrac{x-4}{x-2}=-\dfrac{3}{4}\)
\(\Leftrightarrow4\left(x-4\right)=-3\left(x-2\right)\)
\(\Leftrightarrow4x-16+3x-6=0\)
\(\Leftrightarrow7x-22=0\)
\(\Leftrightarrow x=\dfrac{22}{7}\)
d, Để P có giá trị nguyên
\(\Leftrightarrow x-4⋮x-2\)
\(\Leftrightarrow\left(x-2\right)-2⋮x-2\)
\(\Leftrightarrow2⋮x-2\Leftrightarrow x-2\inƯ\left(2\right)=\left\{1;-1;2;-2\right\}\)
| \(x-2\) | 1 | -1 | 2 | -2 |
| x | 3 | 1 | 4 | 0 |
e,
\(x^2-9=0\)
\(\Rightarrow x^2=9\Rightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)
Với x=3,có :
\(\dfrac{x-4}{x-2}=\dfrac{3-4}{3-2}=-\dfrac{1}{1}=-1\)
Với x=-3,có :
\(\dfrac{x-4}{x-2}=\dfrac{-3-4}{-3-2}=\dfrac{7}{5}\)
Bài 1:
\(\frac{\frac{x}{x-y}-\frac{y}{x+y}}{\frac{y}{x-y}+\frac{x}{x+y}}=\frac{\frac{x(x+y)-y(x-y)}{(x-y)(x+y)}}{\frac{y(x+y+x(x-y)}{(x-y)(x+y)}}=\frac{\frac{x^2+y^2}{(x-y)(x+y)}}{\frac{x^2+y^2}{(x-y)(x+y)}}=1\)
Bài 2:
a)
ĐKXĐ: \(\left\{\begin{matrix} x-5\neq 0\\ x^2-25\neq 0\\ x+5\neq 0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x-5\neq 0\\ (x-5)(x+5)\neq 0\\ x+5\neq 0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x+5\neq 0\\ x-5\neq 0\end{matrix}\right.\Leftrightarrow x\neq \pm 5\)
b)
\(A=\frac{x(x+5)}{(x+5)(x-5)}-\frac{10x}{(x-5)(x+5)}-\frac{5(x-5)}{(x-5)(x+5)}=\frac{x(x+5)-10x-5(x-5)}{(x-5)(x+5)}\)
\(=\frac{x^2-10x+25}{(x-5)(x+5)}=\frac{(x-5)^2}{(x-5)(x+5)}=\frac{x-5}{x+5}\)
c)
Khi $x=9$ thì $A=\frac{9-5}{9+5}=\frac{2}{7}$
a) ĐKXĐ: \(x\ne\pm\sqrt{5}\)
Ta có: \(\frac{3x-1}{x^2-5}=0\)
\(\Leftrightarrow3x-1=0\)
\(\Leftrightarrow3x=1\)
hay \(x=\frac{1}{3}\)(tm)
Vậy: Khi \(x=\frac{1}{3}\) thì giá trị của biểu thức \(\frac{3x-1}{x^2-5}\) bằng 0
b) ĐKXĐ: \(x\ne-\frac{1}{2}\)
Ta có: \(\frac{x^2-x}{2x+1}=0\)
\(\Leftrightarrow x^2-x=0\)
\(\Leftrightarrow x\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)(tm)
Vậy: Khi x∈{0;1} thì biểu thức \(\frac{x^2-x}{2x+1}\) bằng 0
Bài 4:
x^3+x=0
=>x(x^2+1)=0
=>x=0
Bài 5:
\(3n^3+10n^2-5⋮3n+1\)
\(\Leftrightarrow3n^3+n^2+9n^2-1-4⋮3n+1\)
=>\(3n+1\in\left\{1;-1;2;-2;4;-4\right\}\)
hay \(n\in\left\{0;-\dfrac{2}{3};\dfrac{1}{3};-1;1;-\dfrac{5}{3}\right\}\)
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 \(M=\dfrac{5-1}{2}=\dfrac{4}{2}=2\)
Bài 3:
\(3x^2+2x-1=0\)
\(\Leftrightarrow x^2+\dfrac{2}{3}x-\dfrac{1}{3}=0\)
\(\Leftrightarrow x^2+2.x.\dfrac{1}{3}+\dfrac{1}{9}-\dfrac{4}{9}=0\)
\(\Leftrightarrow\left(x+\dfrac{1}{3}\right)^2-\dfrac{4}{9}=0\)
\(\Leftrightarrow\left(x+\dfrac{1}{3}\right)^2=\dfrac{4}{9}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{3}=\dfrac{2}{3}\\x+\dfrac{1}{3}=\dfrac{-2}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=-1\end{matrix}\right.\)
Vậy...