AC
1
K
Khách
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B
2
KT
14 tháng 2 2018
ĐKXĐ: \(x\ne-1;\) \(x\ne-3;\)\(x\ne-5;\)\(x\ne-7\)
\(\frac{1}{x^2+4x+3}+\frac{1}{x^2+8x+15}+\frac{1}{x^2+12x+35}=\frac{3}{16}\)
\(\Leftrightarrow\)\(\frac{1}{\left(x+1\right)\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+5\right)}+\frac{1}{\left(x+5\right)\left(x+7\right)}=\frac{3}{16}\)
\(\Leftrightarrow\)\(\frac{1}{2}\left(\frac{1}{x+1}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+7}\right)=\frac{3}{16}\)
\(\Leftrightarrow\)\(\frac{1}{x+1}-\frac{1}{x+7}=\frac{3}{8}\)
\(\Leftrightarrow\)\(\frac{6}{\left(x+1\right)\left(x+7\right)}=\frac{3}{8}\)
\(\Rightarrow\)\(3\left(x+1\right)\left(x+7\right)=48\)
\(\Leftrightarrow\)\(x^2+8x+7=16\)
\(\Leftrightarrow\)\(x^2+8x-9=0\)
\(\Leftrightarrow\)\(\left(x+1\right)\left(x-9\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=-1\left(KTMĐKXĐ\right)\\x=9\left(TMĐKXĐ\right)\end{cases}}\)
Vậy...
14 tháng 2 2018
\(\frac{1}{x^2+4x+3}+\frac{1}{x^2+8x+15}+\frac{1}{x^2+12x+35}=\frac{3}{16}\)
\(\Leftrightarrow\frac{1}{x^2+x+3x+3}+\frac{1}{x^2+3x+5x+15}+\frac{1}{x^2+5x+7x+35}=\frac{3}{16}\)
\(\Leftrightarrow\frac{1}{\left(x+1\right)\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+5\right)}+\frac{1}{\left(x+5\right)\left(x+7\right)}=\frac{3}{16}\)
\(\Leftrightarrow\frac{\left(x+5\right)\left(x+7\right)}{\left(x+1\right)\left(x+3\right)\left(x+5\right)\left(x+7\right)}+\frac{\left(x+1\right)\left(x+7\right)}{\left(x+1\right)\left(x+3\right)\left(x+5\right)\left(x+7\right)}+\frac{\left(x+1\right)\left(x+3\right)}{\left(x+1\right)\left(x+3\right)\left(x+5\right)\left(x+7\right)}\)
\(=\frac{3\left(x+1\right)\left(x+3\right)\left(x+5\right)\left(x+7\right)}{\left(x+1\right)\left(x+3\right)\left(x+5\right)\left(x+7\right)}\)
Mẫu của mỗi phân thức bằng nhau nên => tử của mỗi phân thức cũng phải bằng nhau
=> Đến đây thì dễ rồi, bạn giải ra tìm x
8 tháng 9 2021
\(a,2\left(x^3-1\right)-2x^2\left(x+2x^4\right)+x\left(4x^5+4\right)=6\\ \Leftrightarrow2x^3-2-2x^3-4x^6+4x^6+4x-6=0\\ \Leftrightarrow4x-8=0\\ \Leftrightarrow x=2\\ b,\left(2x\right)^2\left(4x-2\right)-\left(x^3-8x^3\right)=15\\ \Leftrightarrow4x^2\left(4x-2\right)+7x^3-15=0\\ \Leftrightarrow16x^3-8x^2+7x^3-15=0\\ \Leftrightarrow23x^3-8x^2-15=0\\ \Leftrightarrow23x^3-23x^2+15x^2-15x+15x-15=0\\ \Leftrightarrow\left(x-1\right)\left(23x^2+15x-15\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x\in\varnothing\left(23x^2+15x-15>0\right)\end{matrix}\right.\)
8 tháng 9 2021
Bài 1:
a: Ta có: \(2\left(x^3-1\right)-2x^2\left(2x^4+x\right)+x\left(4x^5+4\right)=6\)
\(\Leftrightarrow2x^3-2-4x^6-2x^3+4x^6+4x=6\)
\(\Leftrightarrow4x=8\)
hay x=2
b: Ta có: \(\left(2x\right)^2\cdot\left(4x-2\right)-\left(x^3-8x^3\right)=15\)
\(\Leftrightarrow4x^2\left(4x-2\right)-x^3+8x^3=15\)
\(\Leftrightarrow16x^3-8x^2+7x^3=15\)
\(\Leftrightarrow23x^3-8x^2-15=0\)
\(\Leftrightarrow23x^3-23x^2+15x^2-15=0\)
\(\Leftrightarrow23x^2\left(x-1\right)+15\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(23X^2+15x+15\right)=0\)
\(\Leftrightarrow x-1=0\)
hay x=1
DN
1
13 tháng 9 2021
\(a,=\left(x+4\right)^2\\ b,=\left(x-6\right)^2\\ c,=-\left(4x^2-4x+1\right)=-\left(2x-1\right)^2\\ d,=\left(x-1\right)^3\)
NG
1
27 tháng 3 2020
https://olm.vn/hoi-dap/detail/64195114200.html
Bn dưới trl r!!
Chúc bn hc tốt!!
26 tháng 7 2018
xin lỗi nha, bài đó bằng có một cái 1/5 thôi, tại viết sai
26 tháng 7 2018
ĐK : \(X\ne-1;-3;-7;-9\)
\(\frac{1}{x^2+4x+3}+\frac{1}{x^2+8x+15}+\frac{1}{x^2+12x+35}+\frac{1}{x^2+16x+63}=\frac{1}{5}\)
\(\frac{1}{\left(x+2\right)^2-1}+\frac{1}{\left(x+4\right)^2-1}+\frac{1}{\left(x+6\right)^2-1}+\frac{1}{\left(x-8\right)^2-1}=\frac{1}{5}\)
\(\frac{1}{\left(x+2-1\right)\left(x+2+1\right)}+\frac{1}{\left(x+4-1 \right)\left(x+4+1\right)}+\frac{1}{\left(x+6-1\right)\left(x+6+1\right)}+\frac{1}{\left(x+8-1\right)\left(x+8+1\right)}=\frac{1}{5}\)
\(\frac{1}{\left(x+1\right)\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+5\right)}+\frac{1}{\left(x+5\right)\left(x+7\right)}+\frac{1}{\left(x+7\right)\left(x+9\right)}=\frac{1}{5}\)
\(\frac{1}{2}\cdot\left(\frac{1}{x+1}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+4}+....-\frac{1}{x+9}\right)=\frac{1}{5}\)
\(\frac{1}{2}\cdot\left(\frac{1}{x+1}-\frac{1}{x+9}\right)=\frac{1}{5}\)
\(\frac{1}{x+1}-\frac{1}{x+9}=\frac{1}{5}:\frac{1}{2}=\frac{2}{5}\)
\(\frac{8}{\left(x+1\right)\left(x+9\right)}=\frac{2}{5}\)
\(2\left(x+1\right)\left(x+9\right)=40\)
\(2x^2+20x+18=40\Leftrightarrow x^2+10x+9=20\)
\(\Leftrightarrow x^2+10x-11=0\Leftrightarrow x^2+10x-10-1=0\)
\(\Leftrightarrow\left(x^2-1\right)+\left(10x-10\right)=0\Leftrightarrow\left(x-1\right)\left(x+1\right)+10\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+11\right)=0\)
\(\orbr{\begin{cases}x-1=0\\x++11=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\\x=-11\end{cases}}}\)( Thõa mãn )
Vậy ...............
T
24 tháng 10 2023
Bài 1.
a)
\((x-2)(2x-1)-(2x-3)(x-1)-2\\=2x^2-x-4x+2-(2x^2-2x-3x+3)-2\\=2x^2-5x+2-(2x^2-5x+3)-2\\=2x^2-5x+2-2x^2+5x-3-2\\=(2x^2-2x^2)+(-5x+5x)+(2-3-2)\\=-3\)
b)
\(x(x+3y+1)-2y(x-1)-(y+x+1)x\\=x^2+3xy+x-2xy+2y-xy-x^2-x\\=(x^2-x^2)+(3xy-2xy-xy)+(x-x)+2y\\=2y\)
Bài 2.
a)
\((14x^3+12x^2-14x):2x=(x+2)(3x-4)\\\Leftrightarrow 14x^3:2x+12x^2:2x-14x:2x=3x^2-4x+6x-8\\ \Leftrightarrow 7x^2+6x-7=3x^2+2x-8\\\Leftrightarrow (7x^2-3x^2)+(6x-2x)+(-7+8)=0\\\Leftrightarrow 4x^2+4x+1=0\\\Leftrightarrow (2x)^2+2\cdot 2x\cdot 1+1^2=0\\\Leftrightarrow (2x+1)^2=0\\\Leftrightarrow 2x+1=0\\\Leftrightarrow 2x=-1\\\Leftrightarrow x=\frac{-1}2\)
b)
\((4x-5)(6x+1)-(8x+3)(3x-4)=15\\\Leftrightarrow 24x^2+4x-30x-5-(24x^2-32x+9x-12)=15\\\Leftrightarrow 24x^2-26x-5-(24x^2-23x-12)=15\\\Leftrightarrow 24x^2-26x-5-24x^2+23x+12=15\\\Leftrightarrow -3x+7=15\\\Leftrightarrow -3x=8\\\Leftrightarrow x=\frac{-8}3\\Toru\)
17 tháng 8 2020
1) \(8x^3+12x^2+6x+1=\left(2x\right)^3+3.\left(2x\right)^2.1+3.2x.1^2+1^3\)
\(=\left(2x+1\right)^3=\left(2.-2+1\right)^3=-27\)
2) \(8x^3-12x+6x-1=\left(2x\right)^3-3.\left(2x\right)^2.1+3.2x.1^2-1^3\)
\(=\left(2x-1\right)^3=\left(2.-\frac{1}{2}-1\right)^3=-8\)
3)\(\left(1-2x\right)^2-\left(3x+1\right)^2=\left(1-2x+3x+1\right)\left(1-2x-3x-1\right)\)
\(=\left(x+2\right)\left(-5x\right)=\left(-2+2\right).\left(-5.-2\right)=0\)
4) \(\left(2x-3y\right)\left(4x^2+6xy+9y^2\right)=\left(2x-3y\right)\left[\left(2x\right)^2+2x.3y+\left(3y\right)^2\right]\)
\(=\left(2x\right)^3-\left(3y\right)^3=\left(2.-\frac{1}{2}\right)^3-\left(3.-\frac{1}{3}\right)^3=-1-\left(-1\right)=0\)
16 tháng 6 2022
24:
\(\Leftrightarrow\dfrac{1}{\left(x+2\right)\left(x+3\right)}+\dfrac{1}{\left(x+3\right)\left(x+4\right)}+\dfrac{1}{\left(x+4\right)\left(x+5\right)}+\dfrac{1}{\left(x+5\right)\left(x+6\right)}=\dfrac{1}{8}\)
\(\Leftrightarrow\dfrac{1}{x+2}-\dfrac{1}{x+6}=\dfrac{1}{8}\)
\(\Leftrightarrow\left(x+2\right)\left(x+6\right)=8\left(x+6\right)-8\left(x+2\right)\)
\(\Leftrightarrow x^2+8x+12=8x+48-8x-16=32\)
=>(x+10)(x-2)=0
=>x=-10 hoặc x=2
25: \(\Leftrightarrow\dfrac{\left(x+1\right)^2+1}{x+1}+\dfrac{\left(x+4\right)^2+4}{x+4}=\dfrac{\left(x+2\right)^2+2}{x+2}+\dfrac{\left(x+3\right)^2+3}{x+3}\)
\(\Leftrightarrow x+1+\dfrac{1}{x+1}+x+4+\dfrac{4}{x+4}=x+2+\dfrac{2}{x+2}+x+3+\dfrac{3}{x+3}\)
\(\Leftrightarrow\dfrac{1}{x+1}+\dfrac{4}{x+4}=\dfrac{2}{x+2}+\dfrac{3}{x+3}\)
\(\Leftrightarrow x+5=0\)
hay x=-5
MT
1
Em kiểm tra lại số hạng đầu tiên, 1/x^21 chắc chắn là sai đề rồi