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4a) \(\left(a+b\right)^2=a^2+2ab+b^2\)
\(\left(a-b\right)^2+4ab=a^2-2ab+b^2+4ab=a^2+b^2+2ab\)
=> (a+b)^2=(a-b)^2+4ab
- 2x – x2 + 2 – x – (3x2 + 6x + 5x +10) = – 4x2 + 2
- 2x – x2 + 2 – x – 3x2 – 6x – 5x – 10 = – 4x2 + 2 –10x = 10 x = – 1
- 2x2 – 6x + x – 3 = 0
(x – 3)(2x + 1) = 0
x = 3 hay x = -1/2
Bài 1:
Theo đầu bài ta có:
\(a+b+c=0\Rightarrow\hept{\begin{cases}a+b=-c\\a+c=-b\\b+c=-a\end{cases}}\)
Từ đó suy ra:
\(H=a\cdot\left(a+b\right)\cdot\left(a+c\right)\)
\(=a\cdot-c\cdot-b\)
\(=a\cdot b\cdot c\)
\(K=c\cdot\left(c+a\right)\cdot\left(c+b\right)\)
\(=c\cdot-b\cdot-a\)
\(=a\cdot b\cdot c\)
Vậy H = K ( đpcm )
Này bạn, tớ thấy bài 1 đề phải là a + b + c = 0 chứ. Sao lại a + b + b = 0 được
a) Ta dùng hằng đẳng thức: \(\left(a-b\right)^2=\left(a+b\right)^2-4ab\) (1)
Thay a+b=7 và ab=12 vào (1) ta được:
\(\left(a-b\right)^2=7^2-4.12=49-48=1\)
Vậy:.....
b) Ta dùng hằng đẳng thức: \(\left(a+b\right)^2=\left(a-b\right)^2+4ab\) (2)
Thay a-b=6 và ab = 3 vào (2) ta được:
\(\left(a+b\right)^2=6^2+4.3=36+12=48\)
Vậy:....
c) Dùng hằng đẳng thức: \(a^3+b^3=\left(a+b\right)^3-3ab\left(a+b\right)\) (3)
Thay ab = 6 và a+b = -5 vào (3) ta được:
\(a^3+b^3=\left(-5\right)^3-3.6\left(-5\right)=-125-90=-215\)
Vậy......
Bài 1:
a)\(a^2+b^2+c^2=ab+bc+ca\)
\(\Rightarrow2a^2+2b^2+2c^2-2ab-2bc-2ca=0\)
\(\Rightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ca+a^2\right)=0\)
\(\Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
Khi \(a=b=c\)
b)\(\left(a+b+c\right)^2=3\left(a^2+b^2+c^2\right)\)
\(\Rightarrow a^2+b^2+c^2+2ab+2bc+2ca=3a^2+3b^2+3c^2\)
\(\Rightarrow-2a^2-2b^2-2c^2+2ab+2bc+2ca=0\)
\(\Rightarrow-\left(a^2-2ab+b^2\right)-\left(b^2-2bc+c^2\right)-\left(c^2-2ca+a^2\right)=0\)
\(\Rightarrow-\left(a-b\right)^2-\left(b-c\right)^2-\left(c-a\right)^2\le0\)
Khi \(a=b=c\)
c)\(\left(a+b+c\right)^2=3\left(ab+bc+ca\right)\)
\(\Rightarrow a^2+b^2+c^2+2ab+2bc+2ca=3ab+3bc+3ca\)
\(\Rightarrow a^2+b^2+c^2-ab-bc-ca=0\)
\(\Rightarrow2a^2+2b^2+2c^2-2ab-2bc-2ca=0\)
\(\Rightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ca+a^2\right)=0\)
\(\Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
Khi \(a=b=c\)
Bài 2:
Từ \(a+b+c=0\Rightarrow\left(a+b+c\right)^2=0\)
\(\Rightarrow a^2+b^2+c^2+2\left(ab+bc+ca\right)=0\)
\(\Rightarrow-2\left(ab+bc+ca\right)=a^2+b^2+c^2\)
\(\Rightarrow ab+bc+ca=-1\)\(\Rightarrow\left(ab+bc+ca\right)^2=1\)
\(\Rightarrow a^2b^2+b^2c^2+c^2a^2+2\left(a^2bc+b^2ca+c^2ab\right)=1\)
\(\Rightarrow a^2b^2+b^2c^2+c^2a^2+2abc\left(a+b+c\right)=1\)
\(\Rightarrow a^2b^2+b^2c^2+c^2a^2=1\left(vi`....a+b+c=0\right)\)
Khi đó: \(a^2+b^2+c^2=2\Rightarrow\left(a^2+b^2+c^2\right)^2=4\)
\(\Rightarrow a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+c^2a^2\right)=4\)
\(\Rightarrow a^4+b^4+c^4+2=4\Rightarrow a^4+b^4+c^4=2\)
so u cn tk m sl fr u
a2 + b2+ c2 = ab + bc + ca
=> a2 + b2+ c2 -ab - bc - ca = 0
=> 2 ( a2 + b2 + c2 -ab -bc - ca) =0
=> ( a2 - 2ab + b2 ) + ( b2 -2bc + c2 ) + ( c2 - 2ca + a2 ) = 0
<=> ( a-b )2 + ( b -c)2 + ( c- a)2 =0
Do ( a -b)2 \(\ge\)0 ( b-c)2 + \(\ge\)0 ( c -a )2 \(\ge\)0
=> a-b =0 ; b -c = 0 ; c -a = 0
=> a=b ; b = c ; c =a
Vậy a = b = c
Bài 1
\(x^5+x^4+1=x^5+x^4+x^3-x^3-x^2-x+x^2+x+1\)
\(=\left(x^5+x^4+x^3\right)+\left(-x^3-x^2-x\right)+\left(x^2+x+1\right)\)
\(=x^3\left(x^2+x+1\right)-x\left(x^2+x+1\right)+\left(x^2+x+1\right)\)
\(=\left(x^3-x+1\right)\left(x^2+x+1\right)\)
Bài 2
Ta có: \(\left(ax+b\right)\left(x^2+cx+1\right)=ax^3+bx^2+acx^2+bcx+ax+b\)
\(=ax^3+\left(b+ac\right)x^2+\left(bc+a\right)x+b=x^3-3x-2\)
\(\Rightarrow a=1\)
\(\Rightarrow b+ac=0\)
\(\Rightarrow bc+a=-3\)
\(\Rightarrow b=-2\)
Thay giá trị của \(a=1;b=-2\)vào \(b+ac=0\)ta được
\(\Leftrightarrow-2+c=0\Rightarrow c=2\)
Vậy \(a=1;b=-2;c=2\)
Bài 3
Ta có \(\left(x^4-3x^3+2x^2-5x\right)\div\left(x^2-3x+1\right)=x^2+1\left(dư-2x+1\right)\)
\(\Rightarrow b=2x-1\)
Bài 4 (cũng làm tương tự như bài 3 nhé )
Bài 5(bài nãy dễ nên bạn tự làm đi nhé)
Bài 6
\(\left(a+b\right)^2=2\left(a^2+b^2\right)\)
\(\Leftrightarrow a^2+2ab+b^2=2a^2+2b^2\)
\(\Leftrightarrow2a^2+2b^2-a^2-2ab-b^2=0\)
\(\Leftrightarrow a^2-2ab+b^2=0\)
\(\Leftrightarrow\left(a-b\right)^2=0\)\(\Rightarrow a-b=0\Rightarrow a=b\)
Bài 7
\(a^2+b^2+c^2=ab+ac+bc\)
\(\Leftrightarrow2a^2+2b^2+2c^2=2ab+2ac+2bc\)
\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2ac-2bc=0\)
\(\Leftrightarrow a^2+a^2+b^2+b^2+c^2+c^2-2ab-2ac-2bc=0\)
\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(a^2-2ac+c^2\right)=0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2=0\)
\(\Rightarrow a-b=0\Rightarrow a=b\)
\(\Rightarrow b-c=0\Rightarrow b=c\)
\(\Rightarrow a-c=0\Rightarrow a=c\)
Vậy \(a=b=c\)
6) c) x3 - x2 + x = 1
<=> x3 - x2 + x - 1 = 0
<=> (x3 - x2) + (x - 1) = 0
<=> x2 (x - 1) + (x - 1) = 0
<=> (x - 1) (x2 + 1) = 0
=> x - 1 = 0 hoặc x2 + 1 = 0
* x - 1 = 0 => x = 1
* x2 + 1 = 0 => x2 = -1 => x = -1
Vậy x = 1 hoặc x = -1
Bài 5:
a) Đặt \(A=\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(\Rightarrow8A=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(\Rightarrow8A=\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(\Rightarrow8A=\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(\Rightarrow8A=\left(3^{16}-1\right)\left(3^{16}+1\right)\)
\(\Rightarrow8A=3^{32}-1\)
\(\Rightarrow A=\frac{3^{32}-1}{8}\)
b) (7x+6)2 + (5-6x)2 - (10-12x)(7x+6)
=(7x+6)2 + (5-6x)2 - 2(5-6x)(7x+6)
\(=\left(7x+6-5+6x\right)^2\)
\(=\left(13x+1\right)^2\)
1. Ta có: \(\left(a-b\right)^2=\left(a+b\right)^2-4ab\)
Theo đề ta có: \(\left(a-b\right)^2=\left(a+b\right)^2-4ab=5^2-4.2=17\)
Vậy \(\left(a-b\right)^2=17\)
2. Ta có: \(\left(a+b\right)^2=\left(a-b\right)^2+4ab\)
Theo đề ta có: \(\left(a+b\right)^2=\left(a-b\right)^2+4ab=6^2+4.16=100\)
\(\Rightarrow\left[{}\begin{matrix}a+b=10\\a+b=-10\end{matrix}\right.\)
Vậy \(a+b=10\) hoặc \(a+b=-10\)
3. \(a^2+b^2+1=ab+a+b\)
\(\Rightarrow2\left(a^2+b^2+1\right)=2\left(ab+a+b\right)\)
\(\Rightarrow2a^2+2b^2+2=2ab+2a+2b\)
\(\Rightarrow2a^2+2b^2+2-2ab-2a-2b=0\)
\(\Rightarrow\left(a^2-2ab+b^2\right)+\left(a^2-2a+1\right)+\left(b^2-2b+1\right)=0\)
\(\Rightarrow\left(a-b\right)^2+\left(a-1\right)^2+\left(b-1\right)^2=0\) (1)
Vì \(\left(a-b\right)^2\ge0\) \(\forall a;b\)
\(\left(a-1\right)^2\ge0\) \(\forall a\)
\(\left(b-1\right)^2\ge0\) \(\forall b\)
\(\Rightarrow\left(a-b\right)^2+\left(a-1\right)^2+\left(b-1\right)^2\ge0\) \(\forall a;b\) (2)
Từ (1)(2) \(\Rightarrow\left\{{}\begin{matrix}\left(a-b\right)^2=0\\\left(a-1\right)^2=0\\\left(b-1\right)^2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a-b=0\\a-1=0\\b-1=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=b\\a=1\\b=1\end{matrix}\right.\Rightarrow a=b=1\)
Vậy.... đpcm
Chúc bạn học tốt ahihi
Bài 1 : \(a+b=5\)
\(\Leftrightarrow\left(a+b\right)^2=25\)
\(\Leftrightarrow a^2+b^2+2ab=25\)
\(\Leftrightarrow a^2+b^2+2.2=25\)
\(\Leftrightarrow a^2+b^2=21\)
\(\Leftrightarrow a^2+b^2-2ab=21-2ab\)
\(\Leftrightarrow\left(a-b\right)^2=21-2.2\)
\(\Leftrightarrow\left(a-b\right)^2=17\)
Bài 2 :
\(a-b=6\)
\(\Leftrightarrow\left(a-b\right)^2=36\)
\(\Leftrightarrow a^2-2ab+b^2=36\)
\(\Leftrightarrow a^2+b^2-2.16=36\)
\(\Leftrightarrow a^2+b^2=36+32=68\)
\(\Leftrightarrow a^2+b^2+2ab=68+2ab\)
\(\Leftrightarrow\left(a+b\right)^2=68+2.16=100\)
\(\Leftrightarrow\left[{}\begin{matrix}a+b=10\\a+b=-10\end{matrix}\right.\)
Bài 3 :
\(a^2+b^2+1=ab+a+b\)
\(\Leftrightarrow2\left(a^2+b^2+1\right)=2\left(ab+a+b\right)\)
\(\Leftrightarrow2a^2+2b^2+2=2ab+2a+2b\)
\(\Leftrightarrow2a^2+2b^2+2-2ab-2a-2b=0\)
\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(a^2-2a+1\right)+\left(b^2-2b+1\right)=0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(a-1\right)^2+\left(b-1\right)^2=0\)
Do \(\left(a-b\right)^2\ge0;\left(a-1\right)^2\ge0;\left(b-1\right)^2\ge0\)
\(\Rightarrow\left(a-b\right)^2+\left(a-1\right)^2+\left(b-1\right)^2\ge0\)
Dấu " = " xảy ra
\(\Leftrightarrow\left\{{}\begin{matrix}a-b=0\\a-1=0\\b-1=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=b\\a=1\\b=1\end{matrix}\right.\)
Vậy \(a=b=1\)