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Ta có :\(\frac{x^7+x^6+x^5+x^4+x^3+x^2+1}{x^2-1}\)
\(=\frac{x^6\left(x+1\right)+x^4\left(x+1\right)+x^2\left(x+1\right)+\left(x+1\right)}{x^2-1}\)
\(=\frac{\left(x^6+x^4+x^2+1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}\)
\(=\frac{\left(x^6+x^4+x^2+1\right)}{\left(x-1\right)}\)
a: \(\Leftrightarrow7\left(7-3x\right)+12\left(5x+2\right)=84\left(x+13\right)\)
\(\Leftrightarrow49-21x+60x+24=84x+1092\)
\(\Leftrightarrow39x-84x=1092-73\)
=>-45x=1019
hay x=-1019/45
b: \(\Leftrightarrow21\left(x+3\right)-14=4\left(5x+9\right)-7\left(7x-9\right)\)
=>21x+63-14=20x+36-49x+63
=>21x+49=-29x+99
=>50x=50
hay x=1
c: \(\Leftrightarrow7\left(2x+1\right)-3\left(5x+2\right)=21x+63\)
=>14x+7-15x-6-21x-63=0
=>-22x-64=0
hay x=-32/11
d: \(\Leftrightarrow35\left(2x-3\right)-15\left(2x+3\right)=21\left(4x+3\right)-17\cdot105\)
=>70x-105-30x-45=84x+63-1785
=>40x-150-84x+1722=0
=>-44x+1572=0
hay x=393/11
Câu 1:
\(\left(x-2\right)\left(x^2+2x+4\right)+25x=x\left(x+5\right)\left(x-5\right)+8\)
\(\Leftrightarrow x^3-8+25x=x\left(x^2-25\right)+8\)
\(\Leftrightarrow x^3-8+25x=x^3-25x+8\)
\(\Leftrightarrow x^3-8+25x-x^3+25x-8=0\)
\(\Leftrightarrow50x-16=0\)
\(\Leftrightarrow50x=16\)
\(\Leftrightarrow x=\dfrac{8}{25}\)
Câu 2 :
\(\dfrac{x+5}{4}+\dfrac{3+2x}{3}=\dfrac{6x-1}{3}-\dfrac{1-2x}{12}\)
<=> \(\dfrac{3\left(x+5\right)}{12}+\dfrac{4\left(3+2x\right)}{12}=\dfrac{4\left(6x-1\right)}{12}-\dfrac{1-2x}{12}\)
<=>\(\dfrac{3x+15+12+8x}{12}=\dfrac{24x-4-1+2x}{12}\)
<=> 3x + 15 + 12 + 8x = 24x - 4 - 1 +2x
<=> 11x+27 = 26x -5
<=> ( 26x - 5 ) - ( 11x + 27 ) = 0
<=> 15x - 32 = 0
<=> 15x = 32
<=> x = \(\dfrac{32}{15}\)
a: \(A=\dfrac{x-1+2x^2+2x+2-x^2-2x}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{1}{x-1}\)
Lời giải:
ĐKXĐ: $x\neq \pm 4$
PT $\Leftrightarrow \frac{8(x-4)+8(x+4)}{x^2-16}=\frac{25}{3}$
$\Leftrightarrow \frac{16x}{x^2-16}=\frac{25}{3}$
$\Rightarrow 48x=25x^2-400$
$\Leftrightarrow 25x^2-48x-400=0$
$\Leftrightarrow (5x-\frac{24}{5})^2=\frac{10576}{25}$
$\Rightarrow x=\frac{24\pm 4\sqrt{661}}{25}$ (đều thỏa mãn)
a,sửa đề : đk x khác -2; 2
\(x^2+x-2+5x-10=12+x^2-4\)
\(\Leftrightarrow6x-20=0\Leftrightarrow x=\dfrac{10}{3}\left(tm\right)\)
b, \(3x-12+5+5x=105\Leftrightarrow8x=112\Leftrightarrow x=14\)
c, \(3x^2+14x-49=-\left(x^2+2x-15\right)\)
\(\Leftrightarrow4x^2+16x-34=0\Leftrightarrow x=\dfrac{-4\pm5\sqrt{2}}{2}\)
a. ko hỉu đề lắm :v
b.\(\dfrac{x-4}{5}+\dfrac{1+x}{3}=7\)
\(\Leftrightarrow\dfrac{3\left(x-4\right)+5\left(1+x\right)}{15}=\dfrac{105}{15}\)
\(\Leftrightarrow3\left(x-4\right)+5\left(1+x\right)=105\)
\(\Leftrightarrow3x-12+5+5x-105=0\)
\(\Leftrightarrow8x-112=0\)
\(\Leftrightarrow8x=112\)
\(\Leftrightarrow x=14\)
c.\(\left(3x-7\right)\left(x+7\right)=\left(5+x\right)\left(3-x\right)\)
\(\Leftrightarrow3x^2+21x-7x-49=15-5x+3x-x^2\)
\(\Leftrightarrow4x^2+16x-64=0\)
Nghiệm xấu lắm bạn
`(x+19)/3 +(x+13)/5 = (x+7)/7 + (x+1)/9`
`<=> x/3 + 19/3 +x/5 +13/5 = x/7 +1 +x/9 +1/9`
`<=> x/3 +x/5 -x/7 -x/9 = 1+1/9 -19/3 -13/5`
`<=> x (1/3 +1/5 -1/7 -1/9) = -118/45`
`<=> x * 88/315 = -352/45`
`<=> x = -28`
Vậy `S={-28}`
1: \(\Leftrightarrow\left(\dfrac{x+1}{85}+1\right)+\left(\dfrac{x+3}{83}+1\right)=\left(\dfrac{x+5}{81}+1\right)+\left(\dfrac{x+7}{79}+1\right)\)
=>x+86=0
=>x=-86
2: \(\Leftrightarrow\left(\dfrac{x-1}{2015}+1\right)-\left(\dfrac{x+3}{2011}+1\right)=\left(\dfrac{x+7}{2007}+1\right)-\left(\dfrac{x+11}{2003}+1\right)\)
=>x+2014=0
=>x=-2014
3: \(\Leftrightarrow3\left(x+4\right)-2\left(x-3\right)=4x\)
=>4x=3x+12-2x+6
=>4x=x+18
=>3x=18
=>x=6
4: \(\Leftrightarrow15x-5\left(x+1\right)=3\left(2x+1\right)\)
=>15x-5x-5=6x+3
=>10x-5=6x+3
=>4x=8
=>x=2
5: \(\Leftrightarrow2\left(2x-7\right)+5\left(x+11\right)=-40\)
=>4x-14+5x+55=-40
=>9x+41=-40
=>x=-9
a: =>3,6-1,7x=2,3-1,4-4=0,9-4=-3,1
=>1,7x=6,7
hay x=67/17
b: \(\Leftrightarrow30\left(5x+4\right)-15\left(3x+5\right)=24\left(4x+9\right)-40\left(x-9\right)\)
=>150x+120-45x-75=96x+216-40x+360
=>105x+45=56x+576
=>49x=531
hay x=531/49
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