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\(3,\\ a,=a^2+2a+1-a^2+2a-1-3a^2+3=-3a^2+4a+3\\ b,=\left(m^3-m+1-m^2+3\right)^2=\left(m^3-m^2-m+4\right)^2\\ 4,\\ a,\Leftrightarrow25x^2+10x+1-25x^2+9=3\\ \Leftrightarrow10x=-7\Leftrightarrow x=-\dfrac{7}{10}\\ b,\Leftrightarrow-9x^2+30x-25+9x^2+18x+9=30\\ \Leftrightarrow48x=46\Leftrightarrow x=\dfrac{23}{24}\\ c,\Leftrightarrow x^2+8x+16-x^2+1=16\\ \Leftrightarrow8x=-1\Leftrightarrow x=-\dfrac{1}{8}\)
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a vì a+2>5 =>a+2+(-2)>5+(-2)=>a+2>3
b vì a>3 => a+2>3+2 =>a+2>5
c vì m>n =>m-n>n-n=>m-n>0
đ vì m-n=0 =>m-n+n>0+n=>m>n
e vì m<n nên m+(-4)<n+(-4) =>m-4<n-4 (1)
vì -4>-5 => m-4>m-5 (2)
từ (1) và (2) =>m-5<n-4
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Bài 2:
a: \(M=\dfrac{x^2+2x}{2\left(x+5\right)}+\dfrac{x-5}{x}+\dfrac{50-5x}{2x\left(x+5\right)}\)
\(=\dfrac{x^3+2x^2+2\left(x-5\right)\left(x+5\right)+50-5x}{2x\left(x+5\right)}\)
\(=\dfrac{x^3+2x^2+50-5x+2x^2-50}{2x\left(x+5\right)}\)
\(=\dfrac{x^3+4x^2-5x}{2x\left(x+5\right)}=\dfrac{x\left(x^2+4x-5\right)}{2x\left(x+5\right)}=\dfrac{x-1}{2}\)
b: Khi x=3 thì \(M=\dfrac{3-1}{2}=\dfrac{2}{2}=1\)
Khi x=5 thì \(M=\dfrac{5-1}{2}=\dfrac{4}{2}=2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
ta có a=5k+3
Nên a2= (5k+3)2=25k2+30k+9=25k2+30k+5+4=5(5k2+6k+1)+4 chia cho 5 dư 4 (dpcm)
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\(A=-x^2+4x-5\\ A=-\left(x^2-4x+4\right)-1\\ A=-\left(x^2-2\cdot x\cdot2+2^2\right)-1\\ A=-\left(x-2\right)^2-1\\ Do\text{ }\left(x-2\right)^2\ge0\forall x\\ \Leftrightarrow-\left(x-2\right)^2\le0\forall x\\ \Leftrightarrow A=-\left(x-2\right)^2-1\le-1\forall x\\ \text{Dấu }"="\text{ xảy ra khi: }\\ \left(x-2\right)^2=0\\ \Leftrightarrow x-2=0\\ \Leftrightarrow x=2\\ \text{Vậy }A_{\left(Max\right)}=-1\text{ }khi\text{ }x=2\)
\(B=-2x^2-6x+5\\ B=-2x^2-6x-\dfrac{9}{2}+\dfrac{19}{2}\\ B=-\left(2x^2+6x+\dfrac{9}{2}\right)+\dfrac{19}{2}\\ B=-2\left(x^2+3x+\dfrac{9}{4}\right)+\dfrac{19}{2}\\ B=-2\left[x^2+2\cdot x\cdot\dfrac{3}{2}+\left(\dfrac{3}{2}\right)^2\right]+\dfrac{19}{2}\\ B=-2\left(x+\dfrac{3}{2}\right)^2+\dfrac{19}{2}\\ Do\text{ }\left(x+\dfrac{3}{2}\right)^2\ge0\forall x\\ \Leftrightarrow2\left(x+\dfrac{3}{2}\right)^2\ge0\forall x\\ \Leftrightarrow-2\left(x+\dfrac{3}{2}\right)^2\le0\forall x\\ \Leftrightarrow B=-2\left(x+\dfrac{3}{2}\right)^2+\dfrac{19}{2}\le\dfrac{19}{2}\forall x\\ \text{Dấu }"="\text{ xảy }ra\text{ }khi:\\ \left(x+\dfrac{3}{2}\right)^2=0\\ \Leftrightarrow x+\dfrac{3}{2}=0\\ \Leftrightarrow x=-\dfrac{3}{2}\\ \text{Vậy }B_{\left(Max\right)}=\dfrac{19}{2}\text{ }khi\text{ }x=-\dfrac{3}{2}\)
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Lời giải:
(A=-x^2+4x-5)
(A=-left(x^2-4x+5 ight))
(A=-left(x^2-4x+4+1 ight))
(A=-left(x^2-4x+4 ight)-1)
(A=-left(x-2 ight)^2-1le1)
Dấu "=" xảy ra khi: (x=2)
(B=-2x^2-6x+5)
(B=-2left(x^2+3x-dfrac{5}{2} ight))
(B=-2left(x^2+3x-dfrac{19}{4}+dfrac{9}{4} ight))
(B=-2left(x^2+3x+dfrac{9}{4} ight)+dfrac{19}{2})
(B=-2left(x+dfrac{3}{2} ight)^2+dfrac{19}{2}ledfrac{19}{2})
Dấu "=" xảy ra khi : (x=-dfrac{3}{2})
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài làm
a) x4+x3+2x2+x+1=(x4+x3+x2)+(x2+x+1)=x2(x2+x+1)+(x2+x+1)=(x2+x+1)(x2+1)
b)a3+b3+c3-3abc=a3+3ab(a+b)+b3+c3 -(3ab(a+b)+3abc)=(a+b)3+c3-3ab(a+b+c)
=(a+b+c)((a+b)2-(a+b)c+c2)-3ab(a+b+c)=(a+b+c)(a2+2ab+b2-ac-ab+c2-3ab)=(a+b+c)(a2+b2+c2-ab-ac-bc)
c)Đặt x-y=a;y-z=b;z-x=c
a+b+c=x-y-z+z-x=o
đưa về như bài b
d)nhóm 2 hạng tử đầu lại và 2hangj tử sau lại để 2 hạng tử sau ở trong ngoặc sau đó áp dụng hằng đẳng thức dề tính sau đó dặt nhân tử chung
e)x2(y-z)+y2(z-x)+z2(x-y)=x2(y-z)-y2((y-z)+(x-y))+z2(x-y)
=x2(y-z)-y2(y-z)-y2(x-y)+z2(x-y)=(y-z)(x2-y2)-(x-y)(y2-z2)=(y-z)(x2-2y2+xy+xz+yz)
![](https://rs.olm.vn/images/avt/0.png?1311)
Câu 13 : Phương trình m(x-1) =5-(m-1)x vô nghiệm nếu :
A/ m=1/2 B/ m=1/4 C/ m=3/2 D/ m=1