x+3/ x-3 -x-3/x+3 = 36/ x^2- 9
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\(\frac{1.5.18+2.10.36+3.15.54}{1.3.9+2.6.18+3.9.27}=\frac{1.5.18.\left(1+2.2.2+3.3.3\right)}{1.3.9.\left(1+2.2.2+3.3.3\right)}\)
\(=\frac{1.5.18}{1.3.9}=\frac{10}{3}\)
\(d,\Leftrightarrow x^3-6x^2+12x-8-x^3+27+6x^2+12x+6=15\\ \Leftrightarrow24x=-10\Leftrightarrow x=-\dfrac{5}{12}\\ e,\Leftrightarrow x^3-3x^2+3x-1+8-x^3+3x^2+6x=17\\ \Leftrightarrow9x=10\Leftrightarrow x=\dfrac{10}{9}\\ f,\Leftrightarrow9x^2+18x+9-18x=36+x^3-27\\ \Leftrightarrow x^3-9x^2=0\Leftrightarrow x^2\left(x-9\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=9\end{matrix}\right.\)
Ta có:
a) ( 45 – 5 x 9 ) x 1 x 2 x 3 x 4 x 5 x 6 x 7
= (45 – 45) x 1 x 2 x 3 x 4 x 5 x 6 x 7
= 0 x 1 x 2 x 3 x 4 x 5 x 6 x 7
= 0
b) (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10) x (72 – 8 x 8 – 8)
= (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10) x (72 – 64 – 8)
= (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10) x 0
= 0
c) (36 – 4 x 9) : (3 x 5 x 7 x 9 x 11)
= (36 – 36) : (3 x 5 x 7 x 9 x 11)
= 0 : (3 x 5 x 7 x 9 x 11)
= 0
d) (27 – 3 x 9) : 9 x 1 x 3 x 5 x 7
= (27 – 27) : 9 x 1 x 3 x 5 x 7
= 0 : 9 x 1 x 3 x 5 x 7
=0
a) ( 45 – 5 x 9 ) x 1 x 2 x 3 x 4 x 5 x 6 x 7
= 0 x 1 x 2 x 3 x 4 x 5 x 6 x 7
b) (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10) x (72 – 8 x 8 – 8)
= (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10) x 0
c) (36 – 4 x 9) : (3 x 5 x 7 x 9 x 11)
= 0 : (3 x 5 x 7 x 9 x 11)
d) (27 – 3 x 9) : 9 x 1 x 3 x 5 x 7
= 0 : 9 x 1 x 3 x 5 x 7 Nếu đúng thì k cho mình nhé bạn!
\(\left(y-2\right)\left(y-3\right)+\left(y-2\right)-1=0\)
\(\Leftrightarrow\left(y-2\right)\left(y-3\right)+\left(y-3\right)=0\)
\(\Leftrightarrow\left(y-3\right)^2=0\)
\(\Leftrightarrow y=3\)
\(x^3+27+\left(x+3\right)\left(x-9\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^2-3x+9\right)+\left(x+3\right)\left(x-9\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^2-3x+9+x-9\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^2-2x\right)=0\)
\(\Leftrightarrow\left(x+3\right)x\left(x-2\right)=0\)
\(\Leftrightarrow x\in\left\{0;-3;2\right\}\)
b)(x+3)2-(x-4)(x+8)=1
\(\Rightarrow\)x2+6x+9-(x2+8x-4x-32)=1
⇒x2+6x+9-x2-8x+4x+32=1
⇒2x+41=1
\(\Rightarrow\)2x+41-1=0
\(\Rightarrow\)2x+40=0
⇒2x=-40
\(\Rightarrow\)x=\(\dfrac{-40}{2}\)
⇒x=-20
a) Ta có: \(\dfrac{x}{x-3}-\dfrac{6}{x}-\dfrac{9}{x^2-3x}\)
\(=\dfrac{x^2}{x\left(x-3\right)}-\dfrac{6\left(x-3\right)}{x\left(x-3\right)}-\dfrac{9}{x\left(x-3\right)}\)
\(=\dfrac{x^2-6x+18-9}{x\left(x-3\right)}\)
\(=\dfrac{\left(x-3\right)^2}{x\left(x-3\right)}=\dfrac{x-3}{x}\)
b) Ta có: \(\dfrac{7}{x}-\dfrac{x}{x+6}+\dfrac{36}{x^2+6x}\)
\(=\dfrac{7\left(x+6\right)-x^2+36}{x\left(x+6\right)}\)
\(=\dfrac{7x+42-x^2+36}{x\left(x+6\right)}\)
\(=\dfrac{-\left(x^2-7x-78\right)}{x\left(x+6\right)}\)
\(=\dfrac{-\left(x^2-13x+6x-78\right)}{x\left(x+6\right)}\)
\(=\dfrac{-\left[x\left(x-13\right)+6\left(x-13\right)\right]}{x\left(x+6\right)}\)
\(=\dfrac{13-x}{x}\)
c) Ta có: \(\dfrac{6}{x-3}-\dfrac{2x-6}{x^2-9}-\dfrac{4}{x+3}\)
\(=\dfrac{6\left(x+3\right)-2x+6-4\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{6x+18-2x+6-4x+12}{\left(x-3\right)\left(x+3\right)}=\dfrac{36}{\left(x-3\right)\left(x+3\right)}\)
a)
\(\dfrac{x-2}{4}+\dfrac{2x-3}{3}=\dfrac{x-18}{6}\)
`<=> 3x-6+8x-12=2x-36`
`<=> 3x+8x-2x=-36+6+12`
`<=> 9x=-18`
`<=> x=-2`
b)
\(\dfrac{x+3}{x-3}+\dfrac{3-x}{x+3}=\dfrac{36}{x^2-9}\left(x\ne3;x\ne-3\right)\)
suy ra
`(x+3)^2 +(3-x)(x-3)=36`
`<=>x^2 +6x+9+3x-9-x^2 +3x=36`
`<=> x^2 -x^2 +6x+3x+3x+9-9-36=0`
`<=> 12x-36=0`
`<=> 12x=36`
`<=> x=3 (KTMĐK)
x + 3/x - 3 - x - 3/x + 3 = 36/x^2 - 9
<=>(x + 3)^2 - (x - 3)^2/(x - 3)(x + 3) = 36/x^2 - 9
<=>x^2 + 6x + 9 - x^2 + 6x - 9/x^2 - 9 = 36/x^2 - 9
<=>12x = 36 <=> x = 3
đk : x khác 3 ; -3
\(\dfrac{\left(x+3\right)^2-\left(x-3\right)^2}{x^2-9}=\dfrac{36}{x^2-9}\Rightarrow x^2+6x+9-x^2+6x-9=36\)
\(\Leftrightarrow12x=36\Leftrightarrow x=3\)(ktmđk)
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