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[ (x2 +1)5 - 2(x2 +1)4 + 3(x2 +1)3] : (x2 +1)3
= (x2 +1)5 : (x2 +1)3 - 2(x2 +1)4 : (x2 +1)3 + 3(x2 +1)3 : (x2 +1)3
= (x2 +1)2 - 2(x2 +1) + 3
= [(x2 +1)2 - 2(x2 +1) + 1 ] +2
= (x2 +1 -1)2 +2
= x4 +2
Với mọi x thì x4 >= 0
=> x4 + 2 >=2 > 0
Vậy thương của biểu thức luôn dương với mọi x
\(a,\left(x-3\right)^2-4=0\)
\(\Leftrightarrow\left(x-3\right)^2=4\)
\(\Rightarrow x-3=\pm2\)
\(\hept{\begin{cases}x-3=2\Rightarrow x=5\\x-3=-2\Rightarrow x=1\end{cases}}\)
Vậy \(x=5\)hoặc \(x=1\)
\(b,x^2-2x=24\)
\(\Leftrightarrow x^2-2x+1-1=24\)
\(\Leftrightarrow\left(x-1\right)^2=24+1=25\)
\(\Leftrightarrow x-1=\pm5\)
\(\hept{\begin{cases}x-1=5\Rightarrow x=6\\x-1=-5\Rightarrow x=-4\end{cases}}\)
Vậy \(x=6\) hoặc \(x=-4\)
\(c,\left(2x+1\right)^2+\left(x+3\right)^2-5\left(x-7\right)\left(x+7\right)=0\)
\(\Leftrightarrow4x^2+4x+1+x^2+6x+9-5\left(x^2-49\right)=0\)
\(\Leftrightarrow4x^2+4x+1+x^2+6x+9-5x^2+245=0\)
\(\Leftrightarrow10x+255=0\)
\(\Leftrightarrow10x=-255\)
\(\Leftrightarrow x=\frac{-51}{2}\)
\(d,\left(x-3\right)\left(x^2+3x+9\right)+x\left(x+2\right)\left(2-x\right)=1\)
\(\Leftrightarrow x^3-27+x\left(2x-x^2+4-2x\right)=1\)
\(\Leftrightarrow x^3-27-x^3+4x=1\)
\(\Leftrightarrow4x-27=1\)
\(\Leftrightarrow4x=28\)
\(\Leftrightarrow x=7\)
1) \(\left(x^2-5\right)\left(x+3\right)+\left(x+4\right)\left(x-x^2\right)\)
\(=\left(x+3\right).x^2-5\left(x+3\right)+\left(x+4\right)\left(x-1x^2\right)\)
\(=x^3+3x^2-5x-15+\left(x+4\right)\left(x-x^2\right)\)
\(=x^3+3x^2-5x-15-x^3+x^2-4x^2+4x\)
\(=3x^2-5x-15-3x^2+4x\)
\(=-x-15\)
Bài 1:
Ta có: \(\left(x+1\right)^3-\left(x-1\right)^3-6\left(x-1\right)^2=19\)
\(\Leftrightarrow x^3+3x^2+3x+1-\left(x^3-3x^2+3x-1\right)-6\left(x^2-2x+1\right)-19=0\)
\(\Leftrightarrow x^3+3x^2+3x+1-x^3+3x^2-3x+1-6x^2+12x-6-19=0\)
\(\Leftrightarrow12x-23=0\)
\(\Leftrightarrow12x=23\)
hay \(x=\frac{23}{12}\)
Vậy: \(x=\frac{23}{12}\)
Bài 2: Rút gọn biểu thức
Ta có: \(3\cdot\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^{16}-1\right)\left(2^{16}+1\right)\)
\(=2^{32}-1\)
a) \(x^2 +x +1 = x^2 +x +1/4 +3/4 = (x+1/2)^2 +3/4\)
các câu khác dùng phương pháp tương tự
a) x^2 + x +1 = x^2 + x + 1/4 + 3/4 = ( x+ 1/2)^2 + 3/4
Vì (x+1/2)^2 >= 0 => (x+1/2)^2 + 3/4>=3/4 > 0
b) 4x^2 - 2x + 1 = (2x)^2 - 2x + 1/4 + 3/4 = (2x +1/2)^2 + 3/4
Vì (2x +1/2)^2 >=0 => (2x +1/2)^2 + 3/4 >= 3/4 > 0
c) x^4 -3x^2 + 9 = x^4 - 3x^2 + 9/4 + 25/4 = ( x^2+ 3/2)^2 + 9/4
Vì ( x^2+ 3/2)^2 >= 0 => ( x^2+ 3/2)^2 + 9/4 >=9/4 >0
d) x^2 + y^2 -2x-2y + 2xy +1
= ( x^2 + 2xy + y^2) - 2( x+y) +1
= ( x+y)^2 -2(x+y) +1
= (x +y +1)^2 >=0
g) x^2+y^2+2(x-2y)+6
= (x^2 + 2x +1) + (y^2 -4y+4) +1
= ( x+1)^2 + (y-2)^2 +1
Vì (x+1)^2; (y-2)^2 >= 0 => ( x+1)^2 + (y-2)^2 +1>=1>0
a) \(x^4-x^2+3=\left[\left(x^2\right)^2-2\cdot x^2\cdot\frac{1}{2}+\frac{1}{4}\right]+\frac{11}{4}=\left(x^2-\frac{1}{2}\right)^2+\frac{11}{4}>0\)
=>đpcm
b) \(x^2-x+1=\left(x^2-2\cdot x\cdot\frac{1}{2}+\frac{1}{4}\right)+\frac{3}{4}=\left(x-\frac{1}{2}\right)^2+\frac{3}{4}>0\)
=>đpcm
c) \(x^2+x+2=\left(x^2+2\cdot x+\frac{1}{2}+\frac{1}{4}\right)+\frac{7}{4}=\left(x+\frac{1}{2}\right)^2+\frac{7}{4}>0\)
=>đpcm
d) \(\left(x+3\right)\left(x-11\right)+20\)
\(=x^2-11x+3x-33+20\)
\(=x^2-8x-13\)
\(=\left(x^2-8x+16\right)-29=\left(x+4\right)^2-29\)
Xem lại đề
a) \(A=x^2+x+1=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}>0\) với mọi x
b) \(B=x^2-x+1=\left(x-\frac{1}{2}\right)^2+\frac{3}{4}>0\) với mọi x
c) \(x^2+xy+y^2+1=\left(x+\frac{1}{2}y\right)^2+\frac{3}{4}y^2+1>0\) với mọi x,y
d) bạn kiểm tra lại đề câu d) nhé:
\(x^2+4y^2+z^2-2x-6y+8z+15\)
\(=\left(x-1\right)^2+\left(2y-\frac{6}{4}\right)^2+\left(z+4\right)^2-\frac{13}{4}\)
Khai triển theo tam giác pascan ta có:
I= x^4+14x^2+1 (khai triển ra rồi rút gọn bạn nhé)
I≥1 vì x^4≥0 và 14x^2≥0
hay I luôn dương