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![](https://rs.olm.vn/images/avt/0.png?1311)
Trả lời:
\(A=\frac{5}{x^2-x+1}=\frac{5}{x^2-2.x.\frac{1}{2}+\left(\frac{1}{2}\right)^2+\frac{3}{4}}=\frac{5}{\left(x-\frac{1}{2}\right)^2+\frac{3}{4}}\)
Ta có: \(\left(x-\frac{1}{2}\right)^2\ge0\forall x\)
\(\Leftrightarrow\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\forall x\)
\(\Leftrightarrow\frac{5}{\left(x-\frac{1}{2}\right)^2+\frac{3}{4}}\le\frac{5}{\frac{3}{4}}\forall x\)
\(\Leftrightarrow A\le\frac{20}{3}\forall x\)
Dấu "=" xảy ra khi \(x-\frac{1}{2}=0\Leftrightarrow x=\frac{1}{2}\)
Vậy GTLN của A bằng 20/3 khi x = 1/2
![](https://rs.olm.vn/images/avt/0.png?1311)
a) (7x - 8)(7x + 8) - 10(2x + 3)2 + 5x(3x - 2)2 - 4x(x - 5)2
= 49x2 - 64 - 10(4x2 + 12x + 9) + 5x(9x2 - 12x + 4) - 4x(x2 - 10x + 25)
= 49x2 - 64 - 40x2 - 120x - 90 + 45x3 - 60x2 + 20x - 4x3 + 40x - 100x
= 41x3 - 51x2 - 160x - 154
b) (x2 - 3)(x2 + 3) - 5x2(x + 1)2 - (x2 - 3x)(x2 - 2x) + 4x(x + 2)2
= x4 - 9 - 5x2(x2 + 2x + 1) - x4 + 5x3 - 6x2 + 4x(x2 + 4x + 4)
= 5x3 - 6x2 - 5x4 - 10x3 - 5x2 + 4x3 + 16x2 + 16x - 9
= -5x4 - x3 + 5x2 + 16x - 9
Trả lời:
a , ( 7x - 8 ) ( 7x + 8 ) - 10 ( 2x + 3 )2 + 5x ( 3x - 2 )2 - 4x ( x - 5 )2
= 49x2 - 64 - 10 ( 4x2 + 12x + 9 ) + 5x ( 9x2 - 12x + 4 ) - 4x ( x2 - 10x + 25 )
= 49x2 - 64 - 40x2 + 120x - 90 + 45x3 - 60x2 + 20x - 4x3 + 40x2 - 100x
= 41x3 - 11x2 + 40x - 154
b , ( x2 - 3 ) ( x2 + 3 ) - 5x2 ( x + 1 )2 - ( x2 - 3x ) ( x2 - 2x ) + 4x ( x + 2 )2
= x4 - 9 - 5x2 ( x2 + 2x + 1 ) - ( x4 - 2x3 - 3x3 + 6x2 ) + 4x ( x2 + 4x + 4 )
= x4 - 9 - 5x4 - 10x3 - 5x2 - x4 + 2x3 + 3x3 - 6x2 + 4x3 + 16x2 + 16x
= - 5x4 - x3 + 5x2 + 16x - 9
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có M = x2 - 6x + 8 = x2 - 6x + 9 - 1 = (x - 3)2 - 1 \(\ge-1\)
Dấu "=" xảy ra <=> x - 3= 0
<=> x = 3
Vậy Min M = -1 <=> x = 3
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x=b+c-a,y=a+c-b,z=a+b-c\)
\(x^3+y^3+z^3=\left(x+y+z\right)^3-3\left(x+y\right)\left(y+z\right)\left(z+x\right)\)
\(=\left(a+b+c\right)^3-3.2c.2a.2b\)
\(=\left(a+b+c\right)^3-24abc\)
\(\left(a+b+c\right)^3-x^3-y^3-z^3=\left(a+b+c\right)^3-\left[\left(a+b+c\right)^3-24abc\right]=24abc\)
![](https://rs.olm.vn/images/avt/0.png?1311)
sửa lại đề bài \(3.3^{3n-1}\left(6.3^{n+2}+3\right)-2.3^n\left(3^{n+3}-1\right)=405\)
\(3^n\left(6.3.3.3^n+3\right)-2.3^n\left(3.3.3.3^n-1\right)=405\)
\(3^n\left(54.3^n+3\right)-2.3^n\left(27.3^n-1\right)=405\)
\(54.3^{2n}+3.3^n-54.3^{2n}+2.3^n=405\)
\(5.3^n=405\)
\(3^n=81\)
\(< =>n=4\)
vậy n =4
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 3:
a) \(M=\left(x+y\right)^3+2x^2+4xy+2y^2=\left(x+y\right)^3+2\left(x+y\right)^2=7^3+2.7^2=441\)
b) \(N=\left(x-y\right)^3-x^2+2xy-y^2=\left(x-y\right)^3-\left(x^2-2xy+y^2\right)=\left(x-y\right)^3-\left(x-y\right)^2\)
\(=\left(-5\right)^3-\left(-5\right)^2=-150\)
Bài 4:
a) \(M=4x^2+8x+7=4x^2+8x+4+3=4\left(x+1\right)^2+3\ge3\)
Dấu \(=\)khi \(x+1=0\Leftrightarrow x=-1\).
b) \(N=6x-x^2-7=-\left(x^2-6x+9\right)+2=-\left(x-3\right)^2+2\le2\)
Dấu \(=\)khi \(x-3=0\Leftrightarrow x=3\).
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a;b>0< =>a^3+b^3>a^3-b^3\)
\(a^3+b^3=a-b\)
\(a^3-b^3< a-b\)
\(\frac{a^3-b^3}{a-b}< \frac{a-b}{a-b}\)
\(a^2+b^2+ab< 1\)
\(< =>ĐPCM\)
Ta có:\(a^2+b^2+ab< 1\)
\(\Leftrightarrow\left(a-b\right)\left(a^2+b^2+ab\right)< a-b\)
\(\Leftrightarrow a^3-b^3< a-b\)
Theo đề bài :\(a^3+b^3=a-b\)
\(\Rightarrow a^3-b^3< a^3+b^3\)
\(\Rightarrow a^3-b^3-a^3-b^3< 0\)
\(\Rightarrow-2b^3< 0\left(\forall b\right)\)
\(1:a,x^4+2x^3+x^2\)
\(x^2\left(x^2+2x+1\right)\)
\(x^2\left(x+1\right)^2\)
\(\left(x^2+x\right)^2\)
\(b,25-x^2+4xy-4y^2\)
\(5^2-\left(x-2y\right)^2\)
\(\left(5-x+2y\right)\left(5+x-2y\right)\)
\(c,y^2-6y-9\)
\(\left(y-3\right)^2-18\)
\(\left(y-3\right)^2-\sqrt{18}^2=\left(y-3-3\sqrt{2}\right)\left(y-3+3\sqrt{2}\right)\)
\(d,x^2-2x-3\)
\(x^2-2x+1-4\)
\(\left(x-1\right)^2-2^2=\left(x-1-2\right)\left(x-1+2\right)\)
\(\left(x-3\right)\left(x+1\right)\)
\(e,9-x^2+2xy-y^2\)
\(9-\left(x^2-2xy+y^2\right)=9-\left(x-y\right)^2\)
\(\left(3-x+y\right)\left(3+x-y\right)\)
\(f,x^2-xz+y^2-yz+2xy\)
\(\left(x+y\right)^2-xz-yz\)
\(\left(x+y\right)^2-z\left(x+y\right)\)
\(\left(x+y\right)\left(x+y-z\right)\)
\(g,4x^2+4x-3\)
\(4x^2+4x+1-4\)
\(\left(x+1\right)^2-2^2=\left(x+1-2\right)\left(x+1+2\right)\)
\(\left(x-1\right)\left(x+3\right)\)
\(h,x^2-x-12\)
\(x^2-x+\frac{1}{2}-\frac{25}{2}\)
\(\left(x-\frac{1}{2}\right)^2-\left(\frac{5}{\sqrt{2}}\right)^2\)
\(\left(x-\frac{1}{2}-\frac{5}{\sqrt{2}}\right)\left(x-\frac{1}{2}+\frac{5}{\sqrt{2}}\right)\)
\(\left(x-\frac{1+5\sqrt{2}}{2}\right)\left(x-\frac{1-5\sqrt{2}}{2}\right)\)
\(i,4x^4+4x^2y^2-8y^4\)
\(\left(2x^2\right)^2+4x^2y^2+\left(y^2\right)^2-\left(3y^2\right)^2\)
\(\left(2x^2+y^2\right)^2-\left(3y^2\right)^2\)
\(\left(2x^2+y^2-3y^2\right)\left(2x^2+y^2+3y^2\right)=\left(2x^2-2y^2\right)\left(2x^2+4y^2\right)\)
\(k,x^4+4y^4\)
\(\left(x^2\right)^2+4x^2y^2+\left(2y^2\right)^2-4x^2y^2\)
\(\left(x^2+2y^2\right)^2-\left(2xy\right)^2\)
\(\left(x^2+2y^2-2xy\right)\left(x^2+2y^2+2xy\right)\)
\(l,ab^2+b^3\)
\(b^2\left(a+b\right)\)
\(m,a^3-a\)
\(a\left(a^2-1\right)\)
\(a\left(a-1\right)\left(a+1\right)\)
\(n,ab^2c^3+64ab^2\)
\(ab^2\left(c^3+64\right)\)
\(ab^2\left(c^3+4^3\right)\)
\(ab^2\left(c+4\right)\left(c^2+4c+16\right)\)