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13 tháng 9 2021
\(6,\\ a,\\ 1,A=x^2+3x+7=\left(x+\dfrac{3}{2}\right)^2+\dfrac{19}{4}\ge\dfrac{19}{4}\)
Dấu \("="\Leftrightarrow x=-\dfrac{3}{2}\)
\(2,B=\left(x-2\right)\left(x-5\right)\left(x^2-7x+10\right)=\left(x-2\right)^2\left(x-5\right)^2\ge0\)
Dấu \("="\Leftrightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\)
\(b,\\ 1,A=11-10x-x^2=-\left(x+5\right)^2+36\le36\)
Dấu \("="\Leftrightarrow x=-5\)
17 tháng 6 2019
a) (x-2)(x+2)(x^2-10)-72=(x^2-4)(x^2-82)
b) x^8+x^6+x^4+x^2+1=x^2 (x^4+x^3+x^2+1+1/x^2)
c)(x+y)^4+x^4+y^4=(x+y)^4+(x+y)^4=2 (x+y)^4
17 tháng 6 2019
a) (x-2)(x+2)(x^2 - 10) -72
= (x^2 - 4)(x^2 - 10) - 72
= x^4 - 4x^2 -10x^2 + 40 - 72
= x^4 - 14x^2 - 32
= x^4 - 16x^2 + 2x^2 - 32
= x^2(x^2 - 16) + 2(x^2 - 16)
= (x^2 - 16)(x^2 + 2)
= (x-4)(x+4)(x^2 + 2)
c) (x+y)4 + x4 + y4
= 2x4 + 4xy3 + 6x2y2 + 4x3y + 2y3
= 2(y4 + 2xy3 + 3x2y2 + 2x3y + x4)
= 2(y2 + xy + y2)2
TN
0
19 tháng 6 2019
\(a,\left(x-2\right)\left(x+2\right)\left(x^2-10\right)-72\)
\(=\left(x^2-4\right)\left(x^2-10\right)-72\)
\(=x^4-14x^2+40-72\)
\(=x^4-14x^2-32\)
\(=x^4-16x^2+2x^2-32\)
\(=\left(x^2-16\right)\left(x^2+2\right)\)
\(=\left(x-4\right)\left(x+4\right)\left(x^2+2\right)\)
\(b,x^8+x^6+x^4+x^2+1\)
\(=x^8+x^7+x^6+x^5+x^4-x^7-x^6-x^5-x^4-x^3+x^6+x^5+x^4+x^3+x^2-x^5-x^4-x^3-x^2-x+x^4+x^3+x^2+x+1\)
\(=x^4\left(x^4+x^3+x^2+x+1\right)-x^3\left(x^4+x^3+x^2+x+1\right)+x^2\left(x^4+x^3+x^2+x+1\right)-x\left(x^4+x^3+x^2+x+1\right)+\left(x^4+x^3+x^2+x+1\right)\)
\(=\left(x^4+x^3+x^2+x+1\right)\left(x^4-x^3+x^2-x+1\right)\)
\(c,\left(x+y\right)^4+x^4+y^4\)
\(=x^4+4xy^3+6x^2y^2+4x^3y+y^4+x^4+y^4\)
\(=2x^4+2y^4+4xy^3+4x^3y+6x^2y^2\)
\(=2\left(x^4+y^4+2xy^3+2x^3y+3x^2y^2\right)\)
\(=2\left(x^2+y^2+xy\right)^2\)
\(d,\left(x+1\right)^4+\left(x^2+x+1\right)^2\)
\(=x^4+4x^3+4x+6x^2+1+x^4+x^2+1+2x^3+2x+2x^2\)
\(=2x^4+6x^3+9x^2+6x+2\)
\(=2x^4+2x^3+x^2+4x^3+4x^2+2x+4x^2+4x+2\)
\(=x^2\left(2x^2+2x+1\right)+2x\left(2x^2+2x+1\right)+2\left(2x^2+2x+1\right)\)
\(=\left(2x^2+2x+1\right)\left(x^2+2x+2\right)\)
G
27 tháng 3 2019
\(y^2+4^x+2y-2^{x+1}+2=0\)
\(\Leftrightarrow\left(y^2+2y+1\right)+\left(4^x-2^{x+1}+1\right)=0\)
\(\Leftrightarrow\left(y+1\right)^2+\left(2^x-1\right)^2=0\Leftrightarrow\hept{\begin{cases}y=-1\\x=0\end{cases}}\)
\(\frac{x^2+4x+6}{x+2}+\frac{x^2+16x+72}{x+8}=\frac{x^2+8x+20}{x+4}+\frac{x^2+12x+42}{x+6}\)
\(\Leftrightarrow\frac{x^2+4x+4+2}{x+2}+\frac{x^2+16x+64+8}{x+8}=\frac{x^2+8x+16+4}{x+4}+\frac{x^2+12x+36+6}{x+6}\)
\(\Leftrightarrow2x+10+\frac{2}{x+2}+\frac{8}{x+8}=2x+10+\frac{4}{x+4}+\frac{6}{x+6}\)
\(\Leftrightarrow\frac{2}{x+2}+\frac{8}{x+8}=\frac{4}{x+4}+\frac{6}{x+6}\)
Tới đây quy đồng làm tiếp nhé
LC
1
TV
19 tháng 3 2018
\(\dfrac{x^2+4x+6}{x+2}+\dfrac{x^2+16x+72}{x+8}=\dfrac{x^2+8x+20}{x+4}+\dfrac{x^2+12x+42}{x+6}\)ĐKXĐ là \(x\ne-2;x\ne-8;x\ne-4;x\ne-6\)
\(\dfrac{x^2+4x+4+2}{x+2}+\dfrac{x^2+16x+64+8}{x+8}=\dfrac{x^2+8x+16+4}{x+4}+\dfrac{x^2+12x+36+6}{x+6}\)\(\Leftrightarrow\dfrac{\left(x+2\right)^2+2}{x+2}+\dfrac{\left(x+8\right)^2+8}{x+8}=\dfrac{\left(x+4\right)^2+4}{x+4}+\dfrac{\left(x+6\right)^2+6}{x+6}\)
\(\Leftrightarrow x+2+\dfrac{2}{x+2}+x+8+\dfrac{8}{x+8}=x+4+\dfrac{4}{x+4}+x+6+\dfrac{6}{x+6}\)
\(\Leftrightarrow\dfrac{2}{x+2}+\dfrac{8}{x+8}=\dfrac{4}{x+4}+\dfrac{6}{x+6}\)
\(\Leftrightarrow\left(\dfrac{2}{x+2}-1\right)+\left(\dfrac{8}{x+8}-1\right)=\left(\dfrac{4}{x+4}-1\right)+\left(\dfrac{6}{x+6}-1\right)\)\(\Leftrightarrow\dfrac{-x}{x+2}+\dfrac{-x}{x+8}=\dfrac{-x}{x+4}+\dfrac{-x}{x+6}\)
\(\Leftrightarrow\dfrac{x}{x+2}+\dfrac{x}{x+8}-\dfrac{x}{x+4}-\dfrac{x}{x+6}=0\)
\(\Leftrightarrow x\left(\dfrac{1}{x+2}+\dfrac{1}{x+8}-\dfrac{1}{x+4}-\dfrac{1}{x+6}\right)=0\)
Do \(\dfrac{1}{x+2}+\dfrac{1}{x+8}-\dfrac{1}{x+4}-\dfrac{1}{x+6}\ne0\)
=> x=0
Vậy ....
dễ mà,nhân tung ra rồi chuyển 72 sang là xong