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HN
1 tháng 1 2018
a) \(\dfrac{x^2+2}{x^3-1}+\dfrac{2}{x^2+x+1}+\dfrac{1}{1-x}\)
\(=\dfrac{x^2+2}{\left(x-1\right)\left(x^2+x+1\right)}+\dfrac{2}{x^2+x+1}-\dfrac{1}{x-1}\)
\(=\dfrac{x^2+2+2\left(x-1\right)-\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{x^2+2+2x-2-x^2-x-1}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{x-1}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{1}{x^2+x+1}\)
b) \(\dfrac{9}{x^3-9x}-\dfrac{-1}{x+3}\)
\(=\dfrac{9}{x\left(x-3\right)\left(x+3\right)}+\dfrac{1}{x+3}\)
\(=\dfrac{9+x\left(x-3\right)}{x\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{9+x^2-3x}{x\left(x-3\right)\left(x+3\right)}\)
c) \(\dfrac{x^3-8}{5x+10}.\dfrac{x^2+4x}{x^2+2x+4}\)
\(=\dfrac{x\left(x-2\right)\left(x^2+2x+4\right)\left(x+4\right)}{5\left(x+2\right)\left(x^2+2x+4\right)}\)
\(=\dfrac{x\left(x-2\right)\left(x+4\right)}{5\left(x+2\right)}\)
d) \(\dfrac{5x+10}{4x-8}.\dfrac{4-2x}{x+2}\)
\(=\dfrac{5\left(x+2\right)}{4\left(x-2\right)}.\dfrac{2\left(2-x\right)}{x+2}\)
\(=-\dfrac{10\left(x+2\right)\left(x-2\right)}{4\left(x-2\right)\left(x+2\right)}\)
\(=-\dfrac{5}{2}\)
e) \(\dfrac{\left(x-13\right)^2}{2x^5}.\dfrac{-3x^2}{x-13}\)
\(=\dfrac{x-13}{2x^3}.\dfrac{-3}{1}\)
\(=\dfrac{-3\left(x-13\right)}{2x^3}\)
g) \(\dfrac{x^2+6x+9}{1-x}.\dfrac{\left(x-1\right)^2}{2\left(x+3\right)^2}\)
\(=-\dfrac{\left(x+3\right)^2}{x-1}.\dfrac{\left(x-1\right)^2}{2\left(x+3\right)^2}\)
\(=-\dfrac{\left(x+3\right)^2\left(x-1\right)^2}{2\left(x-1\right)\left(x+3\right)^2}\)
\(=-\dfrac{x-1}{2}\).
3 tháng 8 2017
a. 3.(x-2)+2.(x-3)=13
x=5
b. (x+1).(2-x)-(3x+5).(x+2)=-4x2+1
x=-9/10
c.x.(5-2x)+2x.(x-1)=13
x=13/3
d. (2x+3)2-(x-1)2=0
x=-2/3
e. x2.(3x-2)-8+12=0
x vô ngiệm
f x2+x=0
x=-1
g. x3-5x=0
x=0
~~~~~~~~~~~ai đi ngang qua nhớ để lại k ~~~~~~~~~~~~~
~~~~~~~~~~~~ Chúc bạn sớm kiếm được nhiều điểm hỏi đáp ~~~~~~~~~~~~~~~~~~~
HT
3 tháng 8 2017
a) \(3\left(x-2\right)+2\left(x-3\right)=1\)\(3\)
\(3x-6+2x-6=13\)
\(5x=13+6+6\)
\(5x=25\)
\(x=25\)
c) \(x\left(5-2x\right)+2x\left(x-1\right)=13\)
\(5x-2x^2+2x^2-2x=13\)
\(3x=13\)
\(x=\frac{13}{3}\)
d) \(\left(2x+3\right)^2-\left(x-1\right)^2=0\)
\(\left(2x+3-x+1\right)\left(2x+3+x-1\right)=0\)
\(\left(x+4\right)\left(3x+2\right)=0\)
\(\orbr{\begin{cases}x+4=0\\3x+2=0\end{cases}}=>\orbr{\begin{cases}x=-4\\x=\frac{-2}{3}\end{cases}}\)
f) \(x^2+x=0\)
\(x\left(x+1\right)=0\)
\(=>\orbr{\begin{cases}x=0\\x+1=0\end{cases}=>\orbr{\begin{cases}x=0\\x=-1\end{cases}}}\)
g) \(x^3-5x=0\)
\(x^2\left(x-5\right)=0\)
\(=>\orbr{\begin{cases}x^2=0\\x-5=0\end{cases}}\)
\(=>\orbr{\begin{cases}x=0\\x=5\end{cases}}\) \(\)
\(\)
HL
3
27 tháng 9 2020
(x + 2)2 - (x - 1)(x + 1) = 13
=> (x2 + 2.x.2 + 22 )- (x2 - 1) = 13 ( dùng hẳng đẳng thức số 1 và số 3)
=> x2 + 4x + 4 - x2 + 1 = 13
=> (x2 - x2) + 4x + 4 + 1 = 13
=> 4x + 4 + 1 = 13
=> 4x + 5 = 13
=> 4x = 8
=> x = 2
Vậy x = 2
(x + 1)3 + x(x - 1) = x3 + 4x2
=> x3 + 3.x2.1 + 3.x.12 + 13 + x2 - x - x3 - 4x2 = 0
=> x3 + 3x2 + 3x + 1 + x2 - x - x3 - 4x2 = 0
=> (x3 - x3) + (3x2 + x2 - 4x2) + (3x - x) + 1 = 0
=> 2x + 1 = 0 => 2x = -1 => x = -1/2
(x + 1)(x + 2) - (x + 3)2 = 24
=> x(x + 2) + 1(x + 2) - (x2 + 2.x.3 + 32) = 24
=> x2 + 2x + x + 2 - (x2 + 6x + 9) = 24
=> x2 + 2x + x + 2 - x2 - 6x - 9 = 24
=> (x2 - x2) + (2x + x - 6x) + (2 - 9) = 24
=> -3x - 7 = 24
=> -3x = 31
=> x = -31/3
(x - 1)(x2 + x + 1) + 2x = x3 + 5
Dựa vào hằng đẳng thức : (A - B)(A2 + AB + B2) = A3 - B3
=> (x - 1)(x2 + x.1 + 12) = x3 - 13 = x3 - 1
=> x3 - 1 + 2x - x3 - 5 = 0
=> (x3 - x3) - 1 + 2x - 5 = 0
=> -1 + 2x - 5 = 0
=> -1 + 2x = 5
=> 2x = 6
=> x = 3
F
30 tháng 9 2020
\(\left(x+2\right)^2-\left(x-1\right)\left(x+1\right)=13\)
\(\left(x^2+4x+4\right)-\left(x^2-1\right)=13\)
\(x^2+4x+4-x^2+1=13\)
\(4x+5=13\)
\(4x=8\)
\(x=2\)
b,\(\left(x+1\right)^3+x\left(x-1\right)=x^3+4x^2\)
\(x^3+3x^2+3x+1+x^2-x-x^3-4x^2=0\)
\(2x+1=0\)
\(2x=-1\)
\(x=-\frac{1}{2}\)
DD
2
DD
0
18 tháng 10 2019
a) \(\left(x-3\right)^2-25x^2=0\)
\(\Leftrightarrow\left(x-3\right)^2-\left(5x\right)^2=0\)
\(\Leftrightarrow\left(x-3-5x\right)\left(x-3+5x\right)=0\)
\(\Leftrightarrow\left(-3-6x\right)\left(6x-3\right)=0\)
\(\Rightarrow\orbr{\begin{cases}-3-6x=0\\6x-3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}-6x=3\\6x=3\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{-1}{2}\\x=\frac{1}{2}\end{cases}}}\)
18 tháng 10 2019
b) \(\frac{x^3-1}{4x}=0\)
\(\Leftrightarrow x^3-1=0\)
\(\Leftrightarrow x^3=1\Leftrightarrow x=1\)
BN
1
TT
30 tháng 9 2020
1, (2x+1)3 - (2x+1)(4x2-2x+1) - 3(2x-1)2 = 15
⇔ \(8x^3+12x^2+6x+1-8x^3-1-3\left(4x^2-4x+1\right)=15\)
⇔ \(12^2+6x-12x^2+12x-3=15\)
⇔ \(18x=18\)
⇔ x = 1
2, x(x-4)(x+4) - (x-5)(x2 +5x+25) = 13
⇔ \(x\left(x^2-16\right)-x^3+125=13\)
⇔ \(x^3-16x-x^3=-\text{112}\)
⇔ \(16x=112\)
⇔ x = 7
NV
Nguyễn Việt Lâm
Giáo viên
11 tháng 7 2020
b/
\(\Leftrightarrow2\left(x^2+x+1\right)^2-7\left(x-1\right)^2=13\left(x-1\right)\left(x^2+x+1\right)\)
Đặt \(\left\{{}\begin{matrix}x^2+x+1=a\\x-1=b\end{matrix}\right.\)
\(\Rightarrow2a^2-7b^2=13ab\)
\(\Leftrightarrow2a^2-13ab-7b^2=0\)
\(\Leftrightarrow\left(a-7b\right)\left(2a+b\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}a-7b=0\\2a+b=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+x+1-7\left(x-1\right)=0\\2\left(x^2+x+1\right)+x-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-6x+8=0\\2x^2+3x+1=0\end{matrix}\right.\) (bấm máy)
NV
Nguyễn Việt Lâm
Giáo viên
11 tháng 7 2020
a/ \(\Leftrightarrow\left(x-4\right)\left(2x+1\right)\left(x-2\right)\left(2x+2\right)-6x^2=0\)
\(\Leftrightarrow\left(2x^2-4-7x\right)\left(2x^2-4-2x\right)-6x^2=0\)
Nhận thấy \(x=0\) không phải nghiệm, pt tương đương:
\(\left(2x-\frac{4}{x}-7\right)\left(2x-\frac{4}{x}-2\right)-6=0\)
Đặt \(2x-\frac{4}{x}-7=t\) ta được:
\(t\left(t+5\right)-6=0\)
\(\Leftrightarrow t^2+5t-6=0\Rightarrow\left[{}\begin{matrix}t=1\\t=-6\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x-\frac{4}{x}-7=1\\2x-\frac{4}{x}-7=-6\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x^2-8x-4=0\\2x^2-x-4=0\end{matrix}\right.\) (bấm máy)
\(\left(x-2\right)^2-\left(x+1\right)\left(x-3\right)=13\)
\(x^2-4x+4-x^2+3x-x+3=13\)
\(-2x+7=13\)
\(-2x=13-7\)
\(-2x=6\)
\(x=-3\)
Ta có
\(\left(x-2\right)^2-\left(x+1\right)\left(x-3\right)=13\)
\(\left(x^2-4x+4\right)-\left(x^2-2x-3\right)=13\)
\(-2x+7=13\)
\(x=-3\)
Vậy x=3