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\(\left(a+b\right).\left(b+c\right).\left(c-a\right)+\left(b+c\right).\left(c+a\right).\left(a-b\right)+\left(c+a\right).\left(a+b\right).\left(b-c\right)\)
\(=\left(a+b\right).\left[\left(b+c\right).\left(c-a\right)+\left(c+a\right).\left(a-b\right)\right]+\left(c+a\right).\left(a+b\right).\left(b-c\right)\)
\(=\left(a+b\right).\left(ac-a^2+bc-ab+a^2-ab+ac-bc\right)+\left(c+a\right).\left(a+b\right).\left(b-c\right)\)
\(=-\left(a+b\right).2a.\left(b-c\right)+\left(c+a\right).\left(a+b\right).\left(b-c\right)\)
\(=\left(a+b\right).\left(b-c\right).\left(-2a+c+a\right)=\left(a+b\right).\left(b-c\right).\left(c-a\right)\)
giai lai:
\(\left(b+c\right).\left[\left(a+b\right).\left(c-a\right)+\left(c+a\right).\left(a-b\right)\right]+\left(c+a\right).\left(a+b\right).\left(b-c\right)\)
\(=-\left(b+c\right).2a.\left(b-c\right)+\left(b-c\right).\left(ac+bc+a^2+ab\right)\)
\(=\left(b-c\right).\left(-2ab-2ac+ac+bc+a^2+ab\right)\)
\(=\left(b-c\right).\left(-ab-ac+bc+a^2\right)\)
\(=\left(b-c\right).\left(a+b\right).\left(a-c\right)\)
\(a,\frac{x+2}{6}-\frac{8x+1}{3}=\frac{2-5x}{2}-6\)
\(\Leftrightarrow\frac{x+2}{6}-\frac{\left(8x+1\right)2}{6}=\frac{\left(2-5x\right)3}{6}-\frac{36}{6}\)
=> x + 2 - 16x - 2 = 6 - 15x - 36
<=> x - 16x + 15x = 6 -36 + 2 - 2
<=> 0x = -30
Phương trình vô ngiệm
b, 11 - ( x + 2) = 3(x + 1)
<=> 11 - x - 2= 3x + 3
<=> -x - 3x = 3 - 11 + 2
<=> -4x = -6
<=> x = \(\frac{3}{2}\)
C, tương tự a
c) ĐKXĐ: x \(\ne\)0 và x \(\ne\)-1
Ta có: \(\frac{x+3}{x+1}+\frac{x+2}{x}=2\)
=> \(x\left(x+3\right)+\left(x+1\right)\left(x+2\right)=2x\left(x+1\right)\)
<=> x2 + 3x + x2 + 3x + 2 = 2x2 + 2x
<=> 2x2 + 6x + 2 - 2x2 - 2x = 0
<=> 4x + 2 = 0
<=> 4x = -2
<=> x = -1/2 (tm)
Vậy S = {-1/2}
Bài 1:
a) A= x2 + 4x + 5
=x2+4x+4+1
=(x+2)2+1\(\ge\)0+1=1
Dấu = khi x+2=0 <=>x=-2
Vậy Amin=1 khi x=-2
b) B= ( x+3 ) ( x-11 ) + 2016
=x2-8x-33+2016
=x2-8x+16+1967
=(x-4)2+1967\(\ge\)0+1967=1967
Dấu = khi x-4=0 <=>x=4
Vậy Bmin=1967 <=>x=4
Bài 2:
a) D= 5 - 8x - x2
=-(x2+8x-5)
=21-x2+8x+16
=21-x2+4x+4x+16
=21-x(x+4)+4(x+4)
=21-(x+4)(x+4)
=21-(x+4)2\(\le\)0+21=21
Dấu = khi x+4=0 <=>x=-4
b)đề sai à
ài 1:
a) A= x2 + 4x + 5
=x2+4x+4+1
=(x+2)2+1$\ge$≥0+1=1
Dấu = khi x+2=0 <=>x=-2
Vậy Amin=1 khi x=-2
b) B= ( x+3 ) ( x-11 ) + 2016
=x2-8x-33+2016
=x2-8x+16+1967
=(x-4)2+1967$\ge$≥0+1967=1967
Dấu = khi x-4=0 <=>x=4
Vậy Bmin=1967 <=>x=4
Bài 2:
a) D= 5 - 8x - x2
=-(x2+8x-5)
=21-x2+8x+16
=21-x2+4x+4x+16
=21-x(x+4)+4(x+4)
=21-(x+4)(x+4)
=21-(x+4)2$\le$≤0+21=21
Dấu = khi x+4=0 <=>x=-4
b)đề sai à
a^2x^2 +(a^2+b^2-c^2)x + b^2 > 0
Δ = (a^2+b^2-c^2)^2 - 4a^2b^2 = (a^2+b^2-c^2 + 2ab)(a^2+b^2-c^2 - 2ab)
= [(a+b)^2 - c^2][a-b)^2 - c^2] = (a+b+c)(a+b-c)(a-b+c)(a-b -c)
(a + b + c) > 0
(a + b - c) > 0
(a - b + c) > 0
(a - b - c) < 0
(tính chất các cạnh tam giác)
=> Δ < 0
=> a^2x^2 +(a^2+b^2-c^2)x + b^2 cùng dấu với a^2 > 0
=> a^2x^2 +(a^2+b^2-c^2)x + b^2 > 0
mình cũng chẳng biết đúng ko nhưng mình nghĩ chắc ai đề
a.\(\left(x^2+x\right)^2+3\left(x^2+x\right)+2=\left(x^2+x\right)^2+2\left(x^2+x\right)+\left(x^2+x+2\right)\)
\(=\left(x^2+x\right)\left(x^2+x+2\right)+\left(x^2+x+2\right)=\left(x^2+x+2\right)\left(x^2+x+1\right)\)
b. \(x\left(x+1\right)\left(x+2\right)\left(x+3\right)+1=\left[x\left(x+3\right)\right]\left[\left(x+1\right)\left(x+2\right)\right]+1\)
\(=\left(x^2+3x\right)\left(x^2+3x+2\right)+1\)(1)
Đặt \(t=x^2+3x\)
(1) \(\Leftrightarrow t\left(t+2\right)+1\)
\(=t^2+2t+1=\left(t+1\right)^2\)(2)
Thay \(t=x^2+3x\)vào (2) t/có:
\(\left(t+1\right)^2=\left(x^2+3x+1\right)^2\)
c. dài lắm mình lười làm, bn bấm thử mạng tìm ik nhớ tíck cho mình nha thanks