tính
\(\frac{1}{x-1}\) \(-\)\(\frac{1}{x+1}\)\(-\)\(\frac{2}{x^2+1}\)\(-\)\(\frac{4}{x^4+1}\)\(-\)\(\frac{8}{x^8+1}\)\(-\)\(\frac{16}{x^{16}+1}\)
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\(\left(5x+3\right)^2-2\left(5x+3\right)\left(x+3\right)+\left(x+3\right)^2\)
Dễ thấy đây là hằng đẳng thức thứ hai với 5x + 3 là A và x + 3 là B
Do đó : \(\left(5x+3\right)^2-2\left(5x+3\right)\left(x+3\right)+\left(x+3\right)^2\)
\(=\left(5x+3-x-3\right)^2\)
\(=\left(4x\right)^2\)
\(=16x^2\)
\(x^2-xy-2x+2y.\)
\(=\left(x^2-xy\right)-\left(2x-2y\right)\)
\(=x\left(x-y\right)-2\left(x-y\right)\)
\(=\left(x-y\right)\left(x-2\right)\)
Hok tốt ạ
x2 - xy - 2x + 2y
=(x2 - 2x) - (xy - 2y)
=x(x - 2) - y(x - 2)
=( x - y )(x - 2)
\(\frac{x^2+y^2}{xy}=\frac{25}{12}\)
\(\Rightarrow12x^2+12y^2=25xy\)
\(\Rightarrow12x^2+12y^2+24xy=49xy\)
\(\Rightarrow12\left(x^2+2xy+y^2\right)=49xy\)
\(\Rightarrow\left(x+y\right)^2=\frac{49xy}{12}\)
\(\Rightarrow x+y=\sqrt{\frac{49xy}{12}}\)
Lại có :\(12\left(x^2-2xy+y^2\right)=xy\)
\(\Rightarrow x-y=\sqrt{\frac{xy}{12}}\)
\(\Rightarrow A=\sqrt{\frac{\frac{xy}{12}}{\frac{49xy}{12}}}\)
\(\Rightarrow A=\sqrt{\frac{1}{49}}=\pm\frac{1}{7}\)
Phạm Tuấn Đạt Chỉ kiến thức lớp 7 là đủ rồi bạn ey!À mà \(\sqrt{\frac{1}{49}}=-\frac{1}{7}???\) không có căn bậc 2 của số âm nha bạn!
\(\frac{x^2+y^2}{xy}=\frac{25}{12}\Leftrightarrow\frac{x^2+y^2}{25}=\frac{xy}{12}\)
Đặt \(\frac{x^2+y^2}{25}=\frac{xy}{12}=k\Rightarrow x^2+y^2=25k;xy=12k\)
\(A^2=\frac{\left(x-y\right)^2}{\left(x+y\right)^2}=\frac{x^2-2xy+y^2}{x^2+2xy+y^2}=\frac{25k-2.12k}{25k+2.12k}=\frac{25k-24k}{25k+24k}=\frac{1k}{49k}=\frac{1}{49}\)
\(\Rightarrow A=\sqrt{\frac{1}{49}}=\frac{1}{7}\)
3(x+3) - x2 + 9
= 3(x +3) - ( x2 - 9 )
=3(x+3) - (x2 - 32)
= 3(x + 3) - (x - 3)(x+3)
= (x+3)(3 - x + 3)
= x(x+3)
Chúc học tốt^^
\(\frac{4}{x-1}-\frac{2}{1-x}-\frac{x}{x-1}\)
\(=\frac{4}{x-1}+\frac{-2}{x-1}-\frac{x}{x-1}\)
\(=\frac{2-x}{x-1}\)
ĐKXĐ: \(x\ne1\)
\(\frac{4}{x-1}-\frac{2}{1-x}-\frac{x}{x-1}\)
\(=\frac{4}{x-1}+\frac{2}{x-1}-\frac{x}{x-1}\)
\(=\frac{4+2-x}{x-1}\)
\(=\frac{6-x}{x-1}\)
\(B=\frac{x^2-6x+14}{x^2-6x+12}=\frac{x^2-6x+12+2}{x^2-6x+12}=1+\frac{2}{x^2-6x+12}\)
ta có: \(x^2-6x+12=x^2-2.3.x+3^2+4=\left(x-3\right)^2+4\ge4\)
để Bmax => \(\left(\frac{2}{x^2-6x+12}\right)max\Rightarrow x^2-6x+12min\)và lớn hơn 0 vì 2>0
mà \(\left(x-3\right)^2+4\) \(\ge\)4
dấu = xảy ra khi x-3=0
=> x=3
Vậy \(MaxB=\frac{3}{2}\)khi x=3
\(\frac{1}{x-1}-\frac{1}{x+1}-\frac{2}{x^2+1}-\frac{4}{x^4+1}-\frac{8}{x^5+1}-\frac{16}{x^{16}+1}\)
\(=\frac{x+1-x+1}{\left(x+1\right)\left(x-1\right)}-\frac{2}{x^2+1}-\frac{4}{x^4+1}-\frac{8}{x^8+1}-\frac{16}{x^{16}+1}\)
\(=\frac{2}{x^2-1}-\frac{2}{x^2+1}-\frac{4}{x^4+1}-\frac{8}{x^8+1}-\frac{16}{x^{16}+1}\)
\(=\frac{2\left(x^2+1\right)-2.\left(x^2-1\right)}{x^2-1}-\frac{4}{x^4+1}-\frac{8}{x^8+1}-\frac{16}{x^{16}+1}\)
\(=\frac{2x^2+2-2x^2+2}{\left(x^2+1\right)\left(x^2-1\right)}-\frac{4}{x^4+1}-\frac{8}{x^8+1}-\frac{16}{x^{16}+1}\)
\(=\frac{4}{x^4-1}-\frac{4}{x^4+1}-\frac{8}{x^8+1}-\frac{16}{x^{16}+1}\)
\(=\frac{4\left(x^4+1\right)-4\left(x^4-1\right)}{\left(x^4-1\right)\left(x^4+1\right)}-\frac{8}{x^8+1}-\frac{16}{x^{16}+1}\)
\(=\frac{8}{x^8-1}-\frac{8}{x^8+1}-\frac{16}{x^{16}+1}\)
\(=\frac{8.\left(x^8+1\right)-8\left(x^8-1\right)}{\left(x^8-1\right)\left(x^8+1\right)}-\frac{16}{x^{16}+1}\)
\(=\frac{16}{x^{16}-1}-\frac{16}{x^{16}+1}\)
\(=\frac{16.\left(x^{16}+1\right)-16.\left(x^{16}-1\right)}{\left(x^{16}-1\right)\left(x^{16}+1\right)}\)
\(=\frac{32}{x^{32}-1}\)