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1. | x + 1| + (y + 2)2 = 0
Mà (y + 2)2 \(\ge\) 0
Đẳng thức khi . y + 2 \(\ge\) 0
y \(\ge\) - 2
. x + 1 = 0
. x = -1
a) Ta có: \(\frac{x+2}{2}-\frac{2x-3}{5}=\frac{10x+13}{10}\)
\(\Leftrightarrow\frac{5\left(x+2\right)}{10}-\frac{2\left(2x-3\right)}{10}-\frac{10x+13}{10}=0\)
Suy ra: \(5x+10-4x+6-10x-13=0\)
\(\Leftrightarrow-9x+3=0\)
\(\Leftrightarrow-9x=-3\)
hay \(x=\frac{1}{3}\)
Vậy: Tập nghiệm \(S=\left\{\frac{1}{3}\right\}\)
b) ĐKXĐ: \(x\notin\left\{2;-2\right\}\)
Ta có: \(\frac{x-1}{x-2}-\frac{5}{x+2}=\frac{x^2}{x^2-4}\)
\(\Leftrightarrow\frac{\left(x-1\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\frac{5\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}-\frac{x^2}{\left(x+2\right)\left(x-2\right)}=0\)
Suy ra: \(x^2+x-2-5x+10-x^2=0\)
\(\Leftrightarrow-4x+8=0\)
\(\Leftrightarrow-4x=-8\)
hay x=2(ktm)
Vậy: Tập nghiệm \(S=\varnothing\)
a) \(\dfrac{10^{12}+5^{11}.2^9-5^{13}.2^8}{4.5^5.10^6}\)
\(=\dfrac{2^{12}.5^{12}+5^{11}.2^9-5^{13}.2^8}{2^2.5^5.2^6.5^6}\)
\(=\dfrac{2^{12}.5^{12}+5^{11}.2^9-5^{13}.2^8}{2^8.5^{11}}\)
\(=\dfrac{\left(2^8.5^{11}\right)\left(2^4.5+2-5^2\right)}{2^8.5^{11}}\)
\(=2^4.5+2-5^2\)
\(=57\)
b) \(\dfrac{\left[5\left(x-y\right)^4-3\left(x-y\right)^3+4\left(x-y\right)^2\right]}{\left(y-x\right)^2}\)
\(=\dfrac{\left(x-y\right)^2\left[5\left(x-y\right)^2-3\left(x-y\right)+4\right]}{\left(y-x\right)^2}\)
\(=\dfrac{\left(x^2+y^2-2xy\right)\left[5\left(x-y\right)^2-3\left(x-y\right)+4\right]}{\left(y^2+x^2-2xy\right)}\)
\(=5\left(x-y\right)^2-3\left(x-y\right)+4\)
c) \(\dfrac{\left(x+y\right)^5-2\left(x+y\right)^4+3\left(x+y\right)^3}{-5\left(x+y\right)^3}\)
\(=\dfrac{\left(x+y\right)^3\left[5\left(x+y\right)^2-2\left(x+y\right)+3\right]}{-5\left(x+y\right)^3}\)
\(=\dfrac{5\left(x+y\right)^2-2\left(x+y\right)+3}{-5}\)
\(\frac{7^4}{13^4}:\left(\frac{7}{13}\right)^2=\frac{7^4}{13^4}:\frac{7^2}{13^2}=\frac{7^4}{13^4}.\frac{13^2}{7^2}=\frac{7^2}{13^2}=\frac{49}{169}\)
a) 17 - 14( x + 1 ) = 13 - 4( x + 1 ) - 5( x - 3 )
<=> 17 - 14x - 14 = 13 - 4x - 4 - 5x + 15
<=> 17 - 14 - 13 + 4 - 15 = -4x - 5x + 14x
<=> -21 = 5x
<=> x = -21/5
b) 7( 4x + 3 ) - 4( x - 1 ) = 15( x + 0, 75 ) + 7
<=> 28x + 21 - 4x + 4 = 15x + 45/4 + 7
<=> 28x - 4x - 15x = 45/4 + 7 - 21 - 4
<=> 9x = -27/4
<=> x = -3/4
c) 3x( x + 1 ) - 2x( x + 2 ) = x2 - 1
<=> 3x2 + 3x - 2x2 - 4x = x2 - 1
<=> 3x2 + 3x - 2x2 - 4x - x2 = -1
<=> -x = -1
<=> x = 1
a, \(17-14\left(x+1\right)=13-4\left(x+1\right)-5\left(x-3\right)\)
\(\Leftrightarrow17-14x-14=13-4x-4-5x+15\)
\(\Leftrightarrow3-14x=24-9x\Leftrightarrow3-14x-24+9x=0\)
\(\Leftrightarrow-21-5x=0\Leftrightarrow5x=-21\Leftrightarrow x=-\frac{21}{5}\)
b, \(7\left(4x+3\right)-4\left(x-1\right)=15\left(x+0,75\right)+7\)
\(\Leftrightarrow28x+21-4x+1=15x+\frac{45}{4}+7\)
\(\Leftrightarrow9x=-\frac{27}{4}\Leftrightarrow x=-\frac{3}{4}\)
c, \(3x\left(x+1\right)-2x\left(x+2\right)=x^2-1\)
\(\Leftrightarrow3x^2+3x-2x^2-4x=x^2-1\)
\(\Leftrightarrow x^2-x=x^2-1\Leftrightarrow x=1\)