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1 + 1 = 2
Cs nên tin ko , nhìu lần ngta bảo thế thì toàn thấy tk sai thui
\(\frac{2}{x-3}+\frac{x-5}{x-1}=1\)
\(ĐKXĐ:x\ne1;x\ne3\)
\(pt\Leftrightarrow\frac{2x-2}{x^2-4x+3}+\frac{x^2-8x+15}{x^2-4x+3}=1\)
\(\Leftrightarrow\frac{x^2-6x+13}{x^2-4x+3}=1\)
\(\Leftrightarrow x^2-6x+13=x^2-4x+3\)
\(\Leftrightarrow-2x+10=0\Leftrightarrow x=-5\left(t/mđkxđ\right)\)
Vậy pt có 1 nghiệm là -5
2/x - 3 + x - 5/x - 1 = 1
2(x - 1) + (x - 5)(x - 3) = (x - 3)(x - 1)
-6x + 13 + x^2 = x^2 - 4x + 3
-6x + 13 = -4x + 3
13 = -4x + 3 + 6x
13 = 2x + 3
13 - 3 = 3x
10 = 2x
5 = x
=> x = 5
Mk lm lại nhé! Nãy nhầm.
`[x-3]/[x+1]+[x+2]/[1-x]+5/[x^2-1]` `ĐK: x \ne +-1`
`=[(x-3)(x-1)-(x+2)(x+1)+5]/[(x-1)(x+1)]`
`=[x^2-x-3x+3-x^2-x-2x-2+5]/[(x-1)(x+1)]`
`=[-7x+6]/[x^2-1]`
1) \(\left(x+\dfrac{1}{3}\right)^3=x^3+3.x^2.\dfrac{1}{3}+3.x.\left(\dfrac{1}{3}\right)^2+\left(\dfrac{1}{3}\right)^3\)
\(=x^3+x^2+\dfrac{x}{3}+\dfrac{1}{27}\)
2) \(\left(2x+y^2\right)^3=\left(2x\right)^3+3.\left(2x\right)^2.y^2+3.2x.\left(y^2\right)^2+\left(y^2\right)^3\)
\(=8x^3+12x^2y^2+6xy^4+y^6\)
3) \(\left(\dfrac{1}{2}x^2+\dfrac{1}{3}y\right)^3=\left(\dfrac{1}{2}x^2\right)^3+3.\left(\dfrac{1}{2}x^2\right)^2.\dfrac{1}{3}y+3.\dfrac{1}{2}x^2.\left(\dfrac{1}{3}y\right)^2+\left(\dfrac{1}{3}y\right)^3\)
\(=\dfrac{1}{8}x^6+\dfrac{1}{4}x^4y+\dfrac{1}{6}x^2y^2+\dfrac{1}{27}y^3\)
4) \(\left(3x^2-2y\right)^3=\left(3x^2\right)^3-3.\left(3x^2\right)^2.2y+3.3x^2.\left(2y\right)^2-\left(2y\right)^3\)
\(=27x^6-54x^4y+36x^2y^2-8y^3\)
5) \(\left(\dfrac{2}{3}x^2-\dfrac{1}{2}y\right)^3=\left(\dfrac{2}{3}x^2\right)^3-3.\left(\dfrac{2}{3}x^2\right)^2.\dfrac{1}{2}y+3.\dfrac{2}{3}x^2.\left(\dfrac{1}{2}y\right)^2-\left(\dfrac{1}{2}y\right)^3\)
\(=\dfrac{8}{27}x^6-\dfrac{1}{3}x^4y+\dfrac{1}{2}x^2y^2-\dfrac{1}{8}y^3\)
6) \(\left(2x+\dfrac{1}{2}\right)^3=\left(2x\right)^3+3.\left(2x\right)^2.\dfrac{1}{2}+3.2x.\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3\)
\(=8x^3+6x^2+\dfrac{3}{2}x+\dfrac{1}{8}\)
7) \(\left(x-3\right)^3=x^3-3.x^2.3+3.x.3^2-3^3\)
\(=x^3-9x^2+27x-27\)
8) \(\left(x+1\right)\left(x^2-x+1\right)\)
\(=\left(x+1\right)\left(x^2-x.1+1^2\right)\)
\(=x^3+1^3\)
\(=x+1\)
9) \(\left(x-3\right)\left(x^2+3x+9\right)\)
\(=\left(x-3\right)\left(x^2+x.3+3^2\right)\)
\(=x^3-3^3\)
\(=x^3-27\)
10) \(\left(x-2\right)\left(x^2+2x+4\right)\)
\(=\left(x-2\right)\left(x^2+x.2+2^2\right)\)
\(=x^3-2^3\)
\(=x^3-8\)
11) \(\left(x+4\right)\left(x^2-4x+16\right)\)
\(=\left(x+4\right)\left(x^2-x.4+4^2\right)\)
\(=x^3+4^3\)
\(=x^3+64\)
12) \(\left(x-3y\right)\left(x^2+3xy+9y^2\right)\)
\(=\left(x-3y\right)\left[x^2+x.3y+\left(3y\right)^2\right]\)
\(=x^3-\left(3y\right)^3\)
\(=x^3-27y^3\)
13) \(\left(x^2-\dfrac{1}{3}\right)\left(x^4+\dfrac{1}{3}x^2+\dfrac{1}{9}\right)\)
\(=\left(x^2-\dfrac{1}{3}\right)\left[\left(x^2\right)^2+x^2.\dfrac{1}{3}+\left(\dfrac{1}{3}\right)^2\right]\)
\(=\left(x^2\right)^3-\left(\dfrac{1}{3}\right)^3\)
\(=x^6-\dfrac{1}{27}\)
14) \(\left(\dfrac{1}{3}x+2y\right)\left(\dfrac{1}{9}x^2-\dfrac{2}{3}xy+4y^2\right)\)
\(=\left(\dfrac{1}{3}x+2y\right)\left[\left(\dfrac{1}{3}x\right)^2-\dfrac{1}{3}x.2y+\left(2y\right)^2\right]\)
\(=\left(\dfrac{1}{3}x\right)^3+\left(2y\right)^3\)
\(=\dfrac{1}{27}x^3+8y^3\)
<=> 4(x^2 + 2x + 1) + 4x^2 - 4x +1 - 8(x^2 - 1) = 11
<=> 4x^2 + 8x + 4 + 4x^2 - 4x +1 - 8x^2 +8 - 11 = 0
<=> 4x + 2 = 0
<=> x = - 1/2
1+1=2 nha bn
=2
hk tốt