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1) \(4x^2+4x+1=\left(2x+1\right)^2\)
2)\(9x^2-24xy+16y^2=\left(3x-4y\right)^2\)
3)\(-x^2+10x-25=-\left(x-5\right)^2\)
4)\(1+12x+36x^2=\left(1+6x\right)^2\)
5) \(\dfrac{x^2}{4}+2xy+4y^2=\left(\dfrac{x}{2}+2y\right)^2\)
6) \(4x^2+4xy+y^2=\left(2x+y\right)^2\)
a) \(4x^2-12x+9=\left(2x\right)^2-2.2x.3+3^2=\left(2x-3\right)^2\)
b) \(4x^2+4x+1=\left(2x\right)^2+2.2x.1+1^2=\left(2x+1\right)^2\)
c) \(1+12x+36x^2=1^2+2.6x.1+\left(6x\right)^2=\left(1+6x\right)^2\)
d) \(9x^2-24xy+16y^2=\left(3x\right)^2-2.3x.4y+\left(4y\right)^2=\left(3x-4y\right)^2\)
f) \(-x^2+10x-25=-\left(x^2-10x+25\right)=-\left(x-5\right)^2\)
g) \(-16a^4b^6-24a^5b^5-9a^6b^4=-\left(16a^4b^6+24a^5b^5+9a^6b^4\right)\)
\(=-\left[\left(4a^2b^3\right)^2+2.4a^2b^3.3a^3b^2+\left(3a^3b^2\right)^2\right]\)
\(=-\left(4a^2b^3+3a^3b^2\right)^2\)
h) \(25x^2-20xy+4y^2=\left(5x\right)^2-2.5x.2y+\left(2y\right)^2\) \(=\left(5x-2y\right)^2\)
i) \(25x^4-10x^2y+y^2=\left(5x^2\right)^2-2.5x^2.y+y^2=\left(5x^2-y\right)^2\)
a) \(4x^2-12x+9\)
\(=\left(2x\right)^2-2.2x.3+3^2\)
\(=\left(2x-3\right)^2\)
b) \(4x^2+4x+1\)
\(=\left(2x\right)^2+2.2x.1+1^2\)
\(=\left(2x+1\right)^2\)
c) \(1+12x+36x^2\)
\(=1^2+2.1.6x+\left(6x\right)^2\)
\(=\left(1+6x\right)^2\)
d) \(9x^2-24xy+16y^2\)
\(=\left(3x\right)^2-2.3x.4y+\left(4y\right)^2\)
\(=\left(3x-4y\right)^2\)
e) \(\frac{x^2}{4}+2xy+4y^2\)
\(=\left(\frac{x}{2}\right)^2+2.\frac{x}{2}.2y+\left(2y\right)^2\)
\(=\left(\frac{x}{2}+2y\right)^2\)
f) \(-x^2+10x-25\)
\(=-\left(x^2-10x+25\right)\)
\(=-\left(x^2-2.5x+5^2\right)\)
\(=-\left(x-5\right)^2\)
g) \(-16a^4b^6-24a^5b^5-9a^6b^4\)
\(=-a^4b^4\left(16b^2+24ab+9a^2\right)\)
\(=-a^4b^4\left[\left(4b\right)^2+2.4b.3a+\left(3a\right)^2\right]\)
\(=-a^4b^4\left(4b+3a\right)^2\)
h) \(25x^2-20xy+4y^2\)
\(=\left(5x\right)^2-2.5x.2y+\left(2y\right)^2\)
\(=\left(5x-2y\right)^2\)
i) \(25x^4-10x^2y+y^2\)
\(=\left(5x^2\right)^2-2.5x^2y+y^2\)
\(=\left(5x^2-y\right)^2\)
1. \(\left(5x-y\right)^2\)
2. \(\left(x^2-2y\right)^2\)
3. \(\left(3x^2-2y\right)^2\)
4. \(\left(9^2-2x^3\right)^2\)
1/ \(25x^2-10xy+y^2=\left(5x\right)^2-2.5xy+y^2=\left(5x-y\right)^2\)
2/\(x^4-4x^2y+4y^2=\left(x^2\right)^2-2.x^2.2y+\left(2y\right)^2=\left(x^2-2y\right)^2\)
3/\(9x^4-12x^2y+4y^2=\left(3x^2\right)^2-2.3x^2.2y+4y^2=\left(3x^2-2y\right)^2\)
4/\(9x^4-12x^5+4x^6=\left(3x^2\right)^2-2.3x^2.2x^3+\left(2x^3\right)^2=\left(3x^2-2x^3\right)^2\)
Giải:
a) \(36-12x+x^2\)
\(=6^2-2.6.x+x^2\)
\(=\left(6-x\right)^2\)
b) \(4x^2+12x+9\)
\(=\left(2x\right)^2+2.3.2x+3^2\)
\(=\left(2x+3\right)^2\)
c) \(-25x^6-y^8+10x^3y^4\)
\(=-\left(25x^6+y^8-10x^3y^4\right)\)
\(=-\left(5x^3-y^4\right)^2\)
d) \(\dfrac{1}{4}x^2-5xy+25y^2\)
\(=\left(\dfrac{1}{2}x\right)^2-2.\dfrac{1}{2}x.5y+\left(5y\right)^2\)
\(=\left(\dfrac{1}{2}x-5y\right)^2\)
Vậy ...
a, 36-12x+x2=62-2.6.x+x2=(6-x)2
b, 4x2+12x+9=(2x)2+2.2x.3+32=(2x+3)2
c, -256-y8+10x3y4=-(256+y8-10x3y4=-(53-y4)2
d, 1/4x2-5xy+25y2=(1/2x)2-2.1/2x.5y+(5y)2=(1/2-5y)2
a) \(36-12x+x^2\) \(=6^2-2.6.x+x^2\)
\(=\left(6-x\right)^2\)
b) \(4x^2+12x+9=\left(2x\right)^2+2.2x.3+3^2\)
\(=\left(2x+3\right)^2\)
c) \(-25x^6-y^8+10x^3y^4=-\left[25x^6-10x^3y^4+y^8\right]\)
\(=-\left[\left(5x^3\right)^2-2.5x^3.y^4+\left(y^4\right)^2\right]\)
\(=-\left(5x^3-y^4\right)^2\)
d) \(\dfrac{1}{4}x^2-5xy+25y^2=\left(\dfrac{1}{2}x\right)^2-2.\dfrac{1}{2}x.5y+\left(5y\right)^2\)
\(=\left(\dfrac{1}{2}x-5y\right)^2\)
Học tốt~~~
a. \(36-12x+x^2=6^2-2.6.x+x^2=\left(6-x\right)^2\)
b. \(4x^2+12x+9=\left(2x\right)^2+2.2x.3+3^2=\left(2x+3\right)^2\)
c: \(=-\left(25x^6-10x^3y^4+y^8\right)\)
\(=-\left(5x^3-y^4\right)^2\)
d: \(=\left(\dfrac{1}{2}x\right)^2-2\cdot\dfrac{1}{2}x\cdot5y+\left(5y\right)^2=\left(\dfrac{1}{2}x-5y\right)^2\)
a)Ta có : 9(a + b)2 - 4(a - 2b)2
= [3(a + b) - 2(a - 2b)].[3(a + b) + 2(a - 2b)]
= (3a + 3b - 2a + 4b)(3a + 3b + 2a - 4b)
= (a + 7b)(5a - b)
\(\left(2x^2-1\right)^2,\left(2x-3\right)^2,\left(x-6\right)^2,\left(5x-1\right)^2,\left(x^2+9\right)^2,\left(2x-5\right)^2,\left(x-2y^2\right)^2,\left(x+5y\right)^2,\left(3y-4x\right)^2\)
Câu hỏi khá dài mình nghĩ lần sau bạn nên chia nhỏ ra nhé
Mình sẽ giải thích 1 số câu khó cho bạn còn lại bạn có thể áp dụng tương tự nhé
1) \(4x^4-4x^2+1=\left(2x^2\right)^2-2.2x^2.1+1^2=\left(2x^2-1\right)^2\)
3) \(36+x^2-12x=x^2-2.x.6+6^2=\left(x-6\right)^2\)
5) \(x^4+81+18x^2=\left(x^2\right)^2+2.x^2.9+9^2=\left(x^2+9\right)^2\)
Chú ý: \(\left(a-b\right)^2=\left(b-a\right)^2\) chứng minh cũng rất dễ
Nên \(\left(x-6\right)^2=\left(6-x\right)^2\) hoàn toàn hợp lí, có 2 cách viết bạn nhé.