Tìm số a,b,c biết 4a2+9b2+16c2 =4a+6b+8c+3
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a,\(5ab-45a^3b\)
=\(5ab\left(1-9a^2\right)\)
=\(5ab\left(1-3a\right)\left(1+3a\right)\)
b,\(3a-6ab+5-10b\)
=\(\left(3a-6ab\right)+\left(5-10b\right)\)
=\(3a\left(1-2b\right)+5\left(1-2b\right)\)
=\(\left(1-2b\right)\left(3a+5\right)\)
c,\(a^2-7ab-2a+14b\)
=\(\left(a^2-7ab\right)-\left(2a-14b\right)\)
=\(a\left(a-7b\right)-2\left(a-7b\right)\)
=\(\left(a-7b\right)\left(a-2\right)\)
d,\(4a^2-8b+4a-8ab\)
=\(\left(4a^2-8ab\right)+\left(4a-8b\right)\)
=\(4a\left(a-2b\right)+4\left(a-2b\right)\)
=\(\left(a-2b\right)\left(4a+4\right)\)
=\(4\left(a-2b\right)\left(a+1\right)\)
e,\(a^2-5a+15b-9b^2\)
=\(\left(a^2-9b^2\right)-\left(5a-15b\right)\)
=\(\left(a-3b\right)\left(a+3b\right)-5\left(a-3b\right)\)
=\(\left(a-3b\right)\left(a+3b-5\right)\)
Chọn C.
Ta có: 4a2 + 9b2 = 13ab hay (2a + 3b) 2 = 25ab
Suy ra:
Suy ra
\(=4\left[a^2-\left(b^2-4bc+4c^2\right)\right]\)
\(=4\left[a^2-\left(b-2c\right)^2\right]\)
\(=4\left(a-b+2c\right)\left(a+b-2c\right)\)
\(4a^2-4b^2+16bc-16c^2\)
\(=4a^2-\left(4b^2-16bc+16c^2\right)\)
\(=\left(2a\right)^2-\left[\left(2b\right)^2-2.2b.4c+\left(4c\right)^2\right]\)
\(=\left(2a\right)^2-\left(2b-4c\right)^2\)
\(=\left(2a+2b-4c\right)\left(2a-2b+4c\right)\)
\(=4\left(a+b-c\right)\left(a-b+c\right)\)
`a)x^4+2x^2y+y^2`
`=(x^2+y)^2`
`b)(2a+b)^2-(2b+a)^2`
`=(2a+b-2b-a)(2a+b+2b+a)`
`=(a-b)(3a+3b)`
`=3(a-b)(a+b)`
`c)8a^3-27b^3-2a(4a^2-9b^2)`
`=(2a-3b)(4a^2+6ab+9b^2)-2a(2a-3b)(2a+3b)`
`=(2a-3b)(4a^2+6ab+9b^2-3a^2-6ab)`
`=9b^2(2a-3b)`
a) Ta có: \(x^4+2x^2y+y^2\)
\(=\left(x^2\right)^2+2\cdot x^2\cdot y+y^2\)
\(=\left(x^2+y\right)^2\)
b) Ta có: \(\left(2a+b\right)^2-\left(2b+a\right)^2\)
\(=\left(2a+b-2b-a\right)\left(2a+b+2b+a\right)\)
\(=\left(a-b\right)\left(3a+3b\right)\)
\(=3\left(a+b\right)\left(a-b\right)\)
a) Ta có: \(A=\left(7-2x\right)\left(7+2x\right)+\left(2x+7\right)^2\)
\(=7-4x^2+4x^2+28x+49\)
\(=28x+56\)
b) Ta có: \(B=\left(4x-5\right)^2-\left(2x-1\right)\left(8x-5\right)\)
\(=16x^2-40x+25-\left(16x^2-10x-8x+5\right)\)
\(=16x^2-40x+25-16x^2+18x-5\)
\(=-22x+20\)
c) Ta có: \(C=\left(5x-3\right)^2-2\left(5x-3\right)\left(5-5x\right)+\left(5x-5\right)^2\)
\(=\left(5x-3\right)^2+2\cdot\left(5x-3\right)\left(5x-5\right)+\left(5x-5\right)^2\)
\(=\left(5x-3+5x-5\right)^2\)
\(=\left(10x-8\right)^2\)
\(=100x^2-160x+64\)
d) Ta có: \(D=\left(2a+3b-c\right)\left(2a-3b+c\right)-\left(4a^2-9b^2-c^2\right)\)
\(=\left[\left(2a+\left(3b-c\right)\right)\left(2a-\left(3b-c\right)\right)\right]-\left(4a^2-9b^2-c^2\right)\)
\(=4a^2-\left(3b-c\right)^2-4a^2+9b^2+c^2\)
\(=-9b^2+6bc-c^2+9b^2+c^2\)
=6bc
d) (8a3 – 27b3) – 2a(4a2 – 9b2)
= (2a – 3b)(4a2 + 6ab + 9b2) – 2a(2a – 3b)(2a + 3b)
= (2a – 3b)(4a2 + 6ab + 9b2 – 4a2 – 6ab) = 9b2(2a – 3b)
4a2+9b2+16c2=4a+6b+8c+3 <=>4a2-4a+1+9b2-6b+1+16c2-8c+1=6 <=> (2a-1)2+(3b-1)2+(4c-1)2=6. hướng dẫn đến đây nhe bạn