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\((2x-1)^2+(x+3)^2-5(x+7)(x-7)=0\)
\(< =>4x^2-4x+1+x^2+6x+9-5\left(x^2-7^2\right)=0\\ < =>4x^2-4x+1+x^2+6x+9-5x^2+245=0\\ < =>2x+255=0\\ < =>2x=-255=>x=\dfrac{-255}{2}\)
Vậy \(x=\dfrac{-255}{2}\)
\(\Rightarrow4x^2-4x+1+x^2+6x+9-5x^2+245=0\)
\(\Rightarrow2x+255=0\Rightarrow2x=-255\Rightarrow x=-\dfrac{255}{2}\)
Sai đề à bn?
Sửa lại đề:
a) (x + 5)2 = (x + 5)(x – 5)
\(\Leftrightarrow\)(x + 5)2 - (x + 5)(x - 5) = 0
\(\Leftrightarrow\)(x + 5)(x - 5 + x + 5) = 0
\(\Leftrightarrow\) (x + 5).10 = 0
\(\Leftrightarrow\) x + 5 = 0
\(\Leftrightarrow\) x = -5
Vậy: x = -5
b, A = (x + 1)(x + 2)(x + 3)(x + 4) – 24
= (x + 1)(x + 4)(x + 2)(x + 3) - 24
= (x2 + 5x + 4)(x2 + 5x + 6) - 24 (*)
Đặt x2 + 5x + 5 = t
Thay x2 + 5x + 5 = t vào (*) ta được:
A = (t - 1)(t + 1) - 24
= t2 - 25
= (t + 5)(t - 5)
= (x2 + 5x + 5 + 5)(x2 + 5x + 5 - 5)
= (x2 + 5x + 10)(x2 + 5x)
= (x2 + 5x + 10).x(x + 5) chia hết (x + 5)(Với x ≠ -5)
Vậy A chia hết (x + 5)(Với x ≠ -5)
Theo nguyên lý Dirichlet, trong 3 số \(x^2;y^2;z^2\) luôn có ít nhất 2 số cùng phía so với 1
Không mất tính tổng quát, giả sử đó là \(x^2\) và \(y^2\)
\(\Rightarrow\left(x^2-1\right)\left(y^2-1\right)\ge0\)
\(\Leftrightarrow x^2y^2+1\ge x^2+y^2\)
\(\Leftrightarrow x^2y^2+5x^2+5y^2+25\ge6x^2+6y^2+24\)
\(\Leftrightarrow\left(x^2+5\right)\left(y^2+5\right)\ge6\left(x^2+y^2+4\right)\)
\(\Rightarrow\left(x^2+5\right)\left(y^2+5\right)\left(z^2+5\right)\ge6\left(x^2+y^2+4\right)\left(z^2+5\right)\)
\(=6\left(x^2+y^2+1+3\right)\left(1+1+z^2+3\right)\)
\(\ge6\left(x+y+z+3\right)^2\)
Dấu "=" xảy ra khi \(x=y=z=1\)
Ta có: \(\left(2x+3\right)\left(x-4\right)+\left(x+5\right)\left(x-2\right)=\left(3x-5\right)\left(x-4\right)\)
\(\Leftrightarrow2x^2-8x+3x-12+x^2-2x+5x-10=3x^2-12x-5x+20\)
\(\Leftrightarrow-2x-22+17x-20=0\)
\(\Leftrightarrow15x=42\)
hay \(x=\dfrac{14}{5}\)
a) (x-2)3+6(x+1)2-x3+12=0
\(\Rightarrow\)x3-6x2+12x-8+6(x2+2x+1)-x3+12=0
\(\Rightarrow\)x3-6x2+12x-8+6x2+12x+6-x3+12=0
\(\Rightarrow\)24x+10=0
\(\Rightarrow\)24x=-10
\(\Rightarrow\)x=\(\dfrac{-10}{24}=\dfrac{-5}{12}\)
b)(x-5)(x+5)-(x+3)2+3(x-2)2=(x+1)2-(x-4)(x+4)+3x2
\(\Rightarrow\)x2-25-(x2+6x+9)+3(x2-4x+4)=x2+2x+1-(x2-16)+3x2
\(\Rightarrow\)x2-25-x2-6x-9+3x2-12x+12=x2+2x+1-x2+16+3x2
\(\Rightarrow\)3x2-18x-22=3x2+2x+17
\(\Rightarrow\)3x2-18x-22-3x2-2x-17=0
\(\Rightarrow\)-20x-39=0
\(\Rightarrow\)-20x=39
\(\Rightarrow\)x=\(-\dfrac{39}{20}\)
a) ( x - 1 )( x2 + x + 1 ) + x( x + 2 )( 2 - x ) = 5
<=> x3 - 1 - x( x + 2 )( x - 2 ) = 5
<=> x3 - 1 - x( x2 - 4 ) = 5
<=> x3 - 1 - x3 + 4x = 5
<=> 4x - 1 = 5
<=> 4x = 6
<=> x = 6/4 = 3/2
b) 5x( x - 3 )2 - 5( x - 1 )3 + 15( x + 4 )( x - 4 ) = 5
<=> 5x( x2 - 6x + 9 ) - 5( x3 - 3x2 + 3x - 1 ) + 15( x2 - 16 ) = 5
<=> 5x3 - 30x2 + 45x - 5x3 + 15x2 - 15x + 5 + 15x2 - 240 = 5
<=> 30x - 235 = 5
<=> 30x = 240
<=> x = 8
a,\(\left(x-1\right)\left(x^2+x+1\right)+x\left(x+2\right)\left(2-x\right)=5\)
\(< =>x^3-1+x\left(4-x^2\right)=5\)
\(< =>x^3-1+4x-x^3=5\)
\(< =>4x-1-5=0< =>4x-6=0< =>x=\frac{3}{2}\)
b, \(5x\left(x-3\right)^2-5\left(x-1\right)^3+15\left(x+4\right)\left(x-4\right)=5\)
\(< =>5x\left(x^2-6x+9\right)-5\left(x^3-3x^2+3x-1\right)+15\left(x^2-16\right)=5\)
\(< =>5x^3-30x^2+45x-5x^3+15x^2-15x+5+15x^2-240=5\)
\(< =>\left(5x^3-5x^3\right)+\left(15x^2+15x^2-30x^2\right)+\left(45x-15x\right)+5-240=5\)
\(< =>30x-240=5-5=0< =>x=\frac{24}{3}=8\)
Ta có : \(\frac{x}{x+5}+\frac{5}{x+5}=\frac{x+5}{x+5}=1\)