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a)
\(\dfrac{1}{x+1}+\dfrac{2}{x^3-x^2-x+1}+\dfrac{3}{x^2-1}=0\) (\(x\ne\pm1\))
\(\Rightarrow\dfrac{\left(x-1\right)^2}{\left(x+1\right)\left(x-1\right)^2}+\dfrac{2}{\left(x+1\right)\left(x-1\right)^2}+\dfrac{3\left(x-1\right)}{\left(x+1\right)\left(x-1\right)^2}=0\)
\(\Rightarrow\dfrac{x^2-2x+1+2+3x-3}{\left(x+1\right)\left(x-1\right)^2}=0\)
\(\Rightarrow\dfrac{x^2+x-2}{\left(x+1\right)\left(x-1\right)^2}=0\)
\(\Rightarrow x^2-x+2=0\)
\(\Rightarrow\left(x-1\right)\left(x+2\right)=0\)
=> Th1 :
x- 1 =0
=> x = 1 ( hư cấu vì không thỏa mãn ĐK )
Th2 :
x+2 = 0
=> x = -2 ( hợp lí )
Vậy nghiệm của phương trình là x = -2
a.
\(\dfrac{1}{x^2-7x+12}+\dfrac{1}{x^2-11x+28}+\dfrac{1}{x^2-19x+84}=\dfrac{1}{4}\\ \Rightarrow\dfrac{1}{\left(x-3\right)\left(x-4\right)}+\dfrac{1}{\left(x-4\right)\left(x-7\right)}+\dfrac{1}{\left(x-7\right)\left(x-12\right)}=\dfrac{1}{4}\\ \Rightarrow\dfrac{1}{x-3}-\dfrac{1}{x-12}=\dfrac{1}{4}\\ \Rightarrow\dfrac{-9}{\left(x-3\right)\left(x-12\right)}=\dfrac{1}{4}\\ \Rightarrow x^2-15x+36=-36\\ \)
Tự giải tiếp
a: \(\Leftrightarrow20x^2-12x+15x+5< 10x\left(2x+1\right)-30\)
\(\Leftrightarrow20x^2+3x+5< 20x^2+10x-30\)
=>3x+5<10x-30
=>-7x<-35
hay x>5
b: \(\Leftrightarrow4\left(5x-20\right)-6\left(2x^2+x\right)>4x\left(1-3x\right)-15x\)
\(\Leftrightarrow20x-80-12x^2-6x>4x-12x^2-15x\)
=>14x-80>-11x
=>25x>80
hay x>16/5
\(a.\Leftrightarrow\frac{5x^2+16}{\left(x+4\right)\left(x-4\right)}=\frac{\left(2x-1\right)\left(x-4\right)+\left(3x-1\right)\left(x+4\right)}{\left(x+4\right)\left(x-4\right)}DKXD:x\ne4;-4\)
\(\Rightarrow5x^2+16=2x^2-8x-x+4+3x^2+12x-x-4\)
\(\Leftrightarrow2x=16\)
\(\Leftrightarrow x=8\)
\(b.\Leftrightarrow\frac{\left(y+1\right)\left(y+2\right)-5\left(y-2\right)}{\left(y-2\right)\left(y+2\right)}=\frac{12+\left(y-2\right)\left(y+2\right)}{\left(y-2\right)\left(y+2\right)}.DKXD:y\ne2;-2\)
\(\Rightarrow y^2+2y+y+2-5y+10=12+y^2-4\)
\(\Leftrightarrow-2y=-4\)
\(\Leftrightarrow y=2\)
a: \(\dfrac{1}{x^2+9x+20}+\dfrac{1}{x^2+11x+30}+\dfrac{1}{x^2+13x+42}=\dfrac{1}{18}\)
\(\Leftrightarrow\dfrac{1}{x+4}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+6}+\dfrac{1}{x+6}-\dfrac{1}{x+7}=\dfrac{1}{18}\)
=>\(\dfrac{x+7-x-4}{\left(x+4\right)\left(x+7\right)}=\dfrac{1}{18}\)
=>(x+4)(x+7)=54
=>x^2+11x+28-54=0
=>(x+13)(x-2)=0
=>x=-13 hoặc x=2
b: \(\Leftrightarrow\dfrac{1}{x+1}-\dfrac{1}{x+2}+\dfrac{1}{x+2}-...+\dfrac{1}{x+4}-\dfrac{1}{x+5}=\dfrac{1}{3}\)
=>\(\dfrac{x+5-x-1}{\left(x+5\right)\left(x+1\right)}=\dfrac{1}{3}\)
=>x^2+6x+5=12
=>x^2+6x-7=0
=>(x+7)(x-1)=0
=>x=-7 hoặc x=1
a: \(\Leftrightarrow-12x-4=8x-2-8-6x\)
=>-12x-4=2x-10
=>-14x=-6
hay x=3/7
b: \(\Leftrightarrow3\left(5x-3\right)-2\left(5x-1\right)=-4\)
=>15x-9-10x+2=-4
=>5x-7=-4
=>5x=3
hay x=3/5(loại)
c: \(\Leftrightarrow x^2-4+3x+3=3+x^2-x-2\)
\(\Leftrightarrow x^2+3x-1=x^2-x+1\)
=>4x=2
hay x=1/2(nhận)
a) \(\dfrac{1}{x^2+5x+4}+\dfrac{1}{x^2+11x+28}+\dfrac{1}{x^2+17x+20}=\dfrac{3}{4x-2}\) \(\left(x\ne-1;-4;-7;-10;\dfrac{1}{2}\right)\)
\(\Leftrightarrow\dfrac{3}{\left(x+1\right)\left(x+4\right)}+\dfrac{3}{\left(x+4\right)\left(x+7\right)}+\dfrac{3}{\left(x+7\right)\left(x+10\right)}=\dfrac{9}{4x-2}\)
\(\Leftrightarrow\dfrac{1}{x+1}-\dfrac{1}{x+10}=\dfrac{9}{4x-2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-4\end{matrix}\right.\)
Mà \(x\ne-4\Rightarrow x=-3\)
Vậy \(S=\left\{-3\right\}\)
b) Gọi số chính phương cần tìm đó là \(\overline{abcd}\) ( \(a,b,c,d\in N\) , \(a\ne0\))
Đặt \(\overline{abcd}=k^2\) \(\left(k\in N,31< k< 100\right)\)
=> Sau khi thêm 1 đơn vị vào chữ số hàng nghìn, thêm 3 đơn vị vào chữ số hàng trăm, thêm 5 đơn vị vào chữ số hàng chục, thêm 3 đơn vì vào chữ số hàng đơn vị thì ta được số \(\overline{\left(a+1\right)\left(b+3\right)\left(c+5\right)\left(d+3\right)}\)
Đặt \(\overline{\left(a+1\right)\left(b+3\right)\left(c+5\right)\left(d+3\right)=m^2}\) \(\left(m\in N,31< m< 100\right)\)
Ta có \(m^2-k^2=1353\Leftrightarrow\left(m+k\right)\left(m-k\right)=1353\)
=> \(\left(m+k\right),\left(m-k\right)\inƯ\left(1353\right)=\left\{1;1353;33;41;123;11;451;3\right\}\)
Mà \(m,k\in N;31< m,k< 100\)
=> 64 < m + k < 200
=> m + k = 123
=> m - k = 11
Ta có : m + k + m - k = 123 + 11 <=> m = 67
=> k = 56 ( TM ) => \(k^2=3136\)
Vậy số chính phương cần tìm là 3136
a) điều kiện xác định : \(x\ne0\)
ta có : \(A=\dfrac{x+1}{x^2+x+1}-\dfrac{x-1}{x^2-x+1}=\dfrac{3}{x\left(x^4+x^2+1\right)}\)
\(\Leftrightarrow A=\dfrac{\left(x+1\right)\left(x^2-x+1\right)-\left(x-1\right)\left(x^2+x+1\right)}{\left(x^2+x+1\right)\left(x^2-x+1\right)}=\dfrac{3}{x\left(x^4+x^2+1\right)}\)
\(\Leftrightarrow A=\dfrac{x^3-x^2+x+x^2-x+1-\left(x^3+x^2+x-x^2-x-1\right)}{x^4-x^3+x^2+x^3-x^2+x+x^2-x+1}=\dfrac{3}{x\left(x^4+x^2+1\right)}\) \(\Leftrightarrow A=\dfrac{x^3-x^2+x+x^2-x+1-x^3-x^2-x+x^2+x+1}{x^4+x^2+1}=\dfrac{3}{x\left(x^4+x^2+1\right)}\) \(\Leftrightarrow A=\dfrac{2}{x^4+x^2+1}=\dfrac{3}{x\left(x^4+x^2+1\right)}\Leftrightarrow\left(x^4+x^2+1\right)A=2=\dfrac{3}{x}\) \(\Leftrightarrow2x=3\Leftrightarrow x=\dfrac{3}{2}\left(tmđk\right)\) vậy \(x=\dfrac{3}{2}\)b) điều kiện : \(x\notin\left\{-4;-5;-6;-7\right\}\)
\(B=\dfrac{1}{x^2+9x+20}+\dfrac{1}{x^2+11x+30}+\dfrac{1}{x^2+13x+42}=\dfrac{1}{18}\)
\(\Leftrightarrow B=\dfrac{1}{\left(x+4\right)\left(x+5\right)}+\dfrac{1}{\left(x+5\right)\left(x+6\right)}+\dfrac{1}{\left(x+6\right)\left(x+7\right)}=\dfrac{1}{18}\)
\(\Leftrightarrow B=\dfrac{\left(x+6\right)\left(x+7\right)+\left(x+4\right)\left(x+7\right)+\left(x+4\right)\left(x+5\right)}{\left(x+4\right)\left(x+5\right)\left(x+6\right)\left(x+7\right)}=\dfrac{1}{18}\)
\(\Leftrightarrow B=\dfrac{x^2+13x+42+x^2+11x+28+x^2+9x+20}{\left(x+4\right)\left(x+5\right)\left(x+6\right)\left(x+7\right)}=\dfrac{1}{18}\)
\(\Leftrightarrow B=\dfrac{3x^2+33x+90}{\left(x+4\right)\left(x+5\right)\left(x+6\right)\left(x+7\right)}=\dfrac{1}{18}\) \(\Leftrightarrow B=\dfrac{3\left(x+5\right)\left(x+6\right)}{\left(x+4\right)\left(x+5\right)\left(x+6\right)\left(x+7\right)}=\dfrac{1}{18}\) \(\Leftrightarrow B=\dfrac{3}{\left(x+4\right)\left(x+7\right)}=\dfrac{1}{18}\) \(\Leftrightarrow54=\left(x+4\right)\left(x+7\right)\)\(\Leftrightarrow54=x^2+11x+28\Leftrightarrow x^2+11x+28-54=0\)
\(\Leftrightarrow x^2+11x-26=0\Leftrightarrow\left(x-2\right)\left(x+13\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+13=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-13\end{matrix}\right.\left(tmđk\right)\)
vậy \(x=2;x=-13\)
ta có x2+5x+4
=x2+x+4x+4
=(x2+x)+(4x+4)
=x(x+1)+4(x+1)
=(x+1)(x+4)
tương tự ta đc
x2+11x+28=(x+4)(x+7)
x2+17x+70=(x+7)(x+10)
x2+23x+130=(x+10)(x+13)
=>\(\dfrac{1}{\left(x+1\right)\left(x+4\right)}+\dfrac{1}{\left(x+4\right)\left(x+7\right)}+\dfrac{1}{\left(x+7\right)\left(x+10\right)}+\dfrac{1}{\left(x+10\right)\left(x+13\right)}=\dfrac{4}{13}\)\(\dfrac{3}{\left(x+1\right)\left(x+4\right)}+\dfrac{3}{\left(x+4\right)\left(x+7\right)}+\dfrac{3}{\left(x+7\right)\left(x+10\right)}+\dfrac{3}{\left(x+10\right)\left(x+11\right)}=\dfrac{4}{13}\)=>\(\dfrac{1}{x+1}-\dfrac{1}{x+4}+\dfrac{1}{x+4}+....+\dfrac{1}{x+13}=\dfrac{4}{13}\)
=>\(\dfrac{1}{x+1}-\dfrac{1}{x+13}=\dfrac{4}{13}\)
=>\(\dfrac{13\left(x+13\right)}{13\left(x+1\right)\left(x+13\right)}-\dfrac{13\left(x+1\right)}{13\left(x+1\right)\left(x+13\right)}=\dfrac{4\left(x+1\right)\left(x+13\right)}{13\left(x+1\right)\left(x+13\right)}\)
=> 13(x+13)-13(x+1)=4(x+1)(x+13)
=> 13[(x+13)-(x+1)]=(4x+4)(x+13)
=>13(x+13-x-1)=4x2+52x+4x+52
=13.12=4x2+56x+52
=>4x2+56x+52=156
=>4x2+56x-104=0