tính :
\(\frac{1}{1-x}+\frac{1}{1+x}+\frac{2}{1+x^2}+\frac{4}{1+x^2}+\frac{4}{1+x^4}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}\)
2y-\(\frac{6xy+2y}{3x+2y}+\frac{2y-9x^2}{3x+2y}\)
Câu 21:
\(\frac{1}{2}\left(\frac{x^{10}}{y^2}+\frac{y^{10}}{x^2}\right)+\frac{1}{4}\left(x^{16}+y^{16}\right)-\left(1+x^2y^2\right)^2\ge x^4y^4+\frac{x^8y^8}{2}-1-2x^2y^2-x^4y^4=\left(x^2y^2-1\right)^2+\frac{1}{2}\left(x^4y^4-1\right)^2-\frac{5}{2}\ge-\frac{5}{2}.\)
Dấu = xảy ra khi x=y=1
Các bạn giỏi Toán vào giúp mình cái:
Tìm giá trị biểu thức: \(P=\frac{3x+2y}{x-2y+4}\) biết\(\frac{2}{x-1}=\frac{3}{2y-3}\)
Thầy giải như sau: \(\frac{2}{x+1}=\frac{3}{2y-3}=\frac{6}{3x+3}=\frac{9}{3x+2y}\)(1)
Mặt khác:\(\frac{2}{x+1}=\frac{3}{2y-3}=\frac{-1}{x+1-2y+3}=\frac{-1}{x-2y+4}\)(2)
Từ (1) và (2) => \(\frac{9}{3x+2y}=\frac{-1}{x-2y+4}\)
=>\(\frac{9}{-1}=\frac{3x+2y}{x-2y+4}=-9\)
Vậy P = -9
Mình thắc mắc chỗ (2) đó các bạn, ở phần \(\frac{-1}{x+1-2y+3}\) đáng lẽ phải là \(\frac{-1}{x+1-2y-3}\) chứ
Giúp mình với
Mình sẽ trình bày rõ hơn ở (2) nha
Ta có:
\(\frac{2}{x+1}=\frac{3}{2y-3}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{2}{x+1}=\frac{3}{2y-3}\) = \(\frac{2-3}{\left(x+1\right)-\left(2y-3\right)}=\frac{-1}{x+1-2y+3}=\frac{-1}{x-2y+4}\)
(Vì trước ngoặc của 2y - 3 là dấu trừ nên khi phá ngoặc thì nó sẽ trở thành dấu cộng.Đây là quy tắc phá ngoặc mà bạn đã được học ở lớp 6 đó)
8,Thực hiện phép tính
a,\(\frac{5x^2-y^2}{xy}-\frac{3x-2y}{y}\)
b,\(\frac{3}{2x+6}-\frac{x-6}{2x^2+6x}\)
c,\(\frac{2x}{x^2+2xy}+\frac{y}{xy-2y^2}+\frac{4}{x^2-4y^2}\)
d,\(\frac{1}{x-y}+\frac{3xy}{y^3-x^3}+\frac{x-y}{x^2+xy+y^2}\)
e,\(\frac{2x+y}{2x^2-xy}+\frac{16x}{y^2-4x^2}+\frac{2x-y}{2x^2+xy}\)
f,\(\frac{1}{1-x}+\frac{1}{1+x}+\frac{2}{1+x^2}+\frac{4}{1+x^4}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}\)
Thực hiện phép tính:
1,\(\frac{1-2x}{2x}+\frac{2x}{2x-1}+\frac{1}{2x-4x^2}\)
2,\(\frac{x^2+2}{x^3-1}+\frac{2}{x^2+x+1}+\frac{1}{1-x}\)
3,\(\frac{x}{x-2y}+\frac{x}{x+2y}+\frac{4xy}{4y^2-x^2}\)
4,\(\frac{2x}{x^2+2xy}+\frac{y}{xy-2y^2}+\frac{4}{x^2-4y^2}\)
5,\(\left(\frac{9}{x^3-9x}+\frac{1}{x+3}\right):\left(\frac{x-3}{x^2+3x}-\frac{x}{3x+9}\right)\)
Làmmmm
1/ \(\frac{1-2x}{2x}+\frac{2x}{2x-1}+\frac{1}{2x-4x^2}\)(ĐKXĐ:x\(\ne0\), x\(\ne\frac{1}{2}\))
= \(\frac{\left(1-2x\right)\left(2x-1\right)}{2x\left(2x-1\right)}+\frac{4x^2}{\left(2x-1\right)2x}-\frac{1}{2x\left(2x-1\right)}\)
\(=\frac{2x-1-4x^2+2x+4x^2-1}{2x\left(2x-1\right)}\)
\(=\frac{4x-2}{2x\left(2x-1\right)}=\frac{2\left(2x-1\right)}{2x\left(2x-1\right)}=\frac{1}{x}\)
KL:..............
2/\(\frac{x^2+2}{x^3-1}+\frac{2}{x^2+x+1}+\frac{1}{1-x}\)(ĐKXĐ : x\(\ne1\))
\(=\frac{x^2+2}{x^3-1}+\frac{2x-2}{x^3-1}-\frac{x^2+x+1}{x^3-1}\)
\(=\frac{x^2+2+2x-2-x^2-x-1}{x^3-1}=\frac{x-1}{x^3-1}=\frac{1}{x^2+x+1}\)
Kl:....................
3/ \(\frac{x}{x-2y}+\frac{x}{x+2y}+\frac{4xy}{4y^2-x^2}\)(x\(\ne\pm2y\))
= \(\frac{x^2+2xy}{x^2-4y^2}+\frac{x^2-2xy}{x^2-4y^2}-\frac{4xy}{x^2-4y^2}=\frac{2x^2-4xy}{x^2-4y^2}=\frac{2x\left(x-2y\right)}{x^2-4y^2}=\frac{2x}{x+2y}\)
Kl:................
giúp mình với
làm phép tính
c) \(\frac{1}{3x-2}-\frac{4}{3x+2}-\frac{3x-6}{4-9x^2}\)
d) x - 2 \(-\frac{x^2-10}{x+2}\)
e)\(\frac{1}{2x-2y}-\frac{1}{2x+2y}+\frac{y}{y^2-x^2}\)
g)\(\frac{4-2x+x^2}{x+2}-2-x\)
i)\(\frac{1}{2x+3}-\frac{1}{2x-3}+\frac{x-2}{2x^2-x-3}\)
c: \(=\dfrac{1}{3x-2}-\dfrac{4}{3x+2}+\dfrac{3x-6}{\left(3x-2\right)\left(3x+2\right)}\)
\(=\dfrac{3x+2-12x+8+3x-6}{\left(3x-2\right)\left(3x+2\right)}\)
\(=\dfrac{-6x+4}{\left(3x-2\right)\left(3x+2\right)}=\dfrac{-2}{3x+2}\)
d: \(=\dfrac{x^2-4-x^2+10}{x+2}=\dfrac{6}{x+2}\)
e: \(=\dfrac{1}{2\left(x-y\right)}-\dfrac{1}{2\left(x+y\right)}-\dfrac{y}{\left(x-y\right)\left(x+y\right)}\)
\(=\dfrac{x+y-x+y-2y}{2\left(x-y\right)\left(x+y\right)}=0\)
Bài 1: Thu gọn
a) \(\frac{1}{5}x^4y^3-3x^4y^3\)
b) \(5x^2y^5-\frac{1}{4}x^2y^5\)
c) \(\frac{1}{7}x^2y^3.\left(-\frac{14}{3}xy^2\right)-\frac{1}{2}xy.\left(x^2y^{\text{4}}\right)\)
d) \(\left(3xy\right)^2.\left(-\frac{1}{2}x^3y^2\right)\)
e) \(-\frac{1}{4}xy^2+\frac{2}{5}x^2y+\frac{1}{2}xy^2-x^2y\)
f) \(\frac{1}{2}x^4y.\left(-\frac{2}{3}x^3y^2\right)-\frac{1}{3}x^7y^3\)
g) \(\frac{1}{2}x^2y.\left(-10x^3yz^2\right).\frac{1}{4}x^5y^3z\)
h) \(4.\left(-\frac{1}{2}x\right)^2-\frac{3}{2}x.\left(-x\right)+\frac{1}{3}x^2\)
i) \(1\frac{2}{3}x^3y.\left(\frac{-1}{2}xy^2\right)^2-\frac{5}{4}.\frac{8}{15}x^3y.\left(-\frac{1}{2}xy^2\right)^2\)
k) \(-\frac{3}{2}xy^2.\left(\frac{3}{4}x^2y\right)^2-\frac{3}{5}xy.\left(-\frac{1}{3}x^4y^3\right)+\left(-x^2y\right)^2.\left(xy\right)^2\)
n) \(-2\frac{1}{5}xy.\left(-5x\right)^2+\frac{3}{4}y.\frac{2}{3}\left(-x^3\right)-\frac{1}{9}.\left(-x\right)^3.\frac{1}{3}y\)
m) \(\left(-\frac{1}{3}xy^2\right)^2.\left(3x^2y\right)^3.\left(-\frac{5}{2}xy^2z^3\right)^{^2}\)
p) \(-2y.\left|2\right|x^4y^5.\left|-\frac{3}{4}\right|x^3y^2z\)
Bài 1:
a) \(\frac{1}{5}x^4y^3-3x^4y^3\)
= \(\left(\frac{1}{5}-3\right)x^4y^3\)
= \(-\frac{14}{5}x^4y^3.\)
b) \(5x^2y^5-\frac{1}{4}x^2y^5\)
= \(\left(5-\frac{1}{4}\right)x^2y^5\)
= \(\frac{19}{4}x^2y^5.\)
Mình chỉ làm 2 câu thôi nhé, bạn đăng nhiều quá.
Chúc bạn học tốt!
bài 1 : thực hiện các phép tính
a. \(\frac{4x-1}{3x^2y}-\frac{7x-1}{3x^2y}\)
b.\(\frac{4x+1}{2}-\frac{3x+2}{3}\)
c.\(\frac{5x^2-y^2}{xy}-\frac{3x-2y}{y}\)
d.\(\frac{3x}{5x+5y}-\frac{x}{10x-10y}\)
e.\(\frac{x+3}{x}-\frac{x}{x-3}+\frac{9}{x^2-3x}\)
f.\(\frac{x+3}{x^2-1}-\frac{1}{x^2+x}\)
g.\(\frac{1}{3x-2}-\frac{4}{3x+2}-\frac{-10x+8}{9x^2-4}\)
h.\(\frac{x+9y}{x^2-9y^2}-\frac{3y}{x^2+3xy}\)
i.\(\frac{3}{2x+6}-\frac{x-6}{2x^2+6x}\)
bài 1: Thực hiện các phép tính
a.\(\frac{4x-1}{3x^2y}-\frac{7x-2}{3x^2y}\)
b.\(\frac{4x+1}{2}-\frac{3x+2}{3}\)
c.\(\frac{5x^2-y^2}{xy}-\frac{3x-2y}{y}\)
d.\(\frac{3x}{5x+5y}-\frac{x}{10x-10y}\)
e. \(\frac{x+3}{x}-\frac{x}{x-3}+\frac{9}{x^2-3x}\)
f..\(\frac{x+3}{x^2-1}-\frac{1}{x^2+x}\)
g. \(\frac{1}{3x-2}-\frac{4}{3x+2}-\frac{-10x+8}{9x^2-4}\)
h.\(\frac{x+9y}{x^2-9y^2}-\frac{3y}{x^2+3xy}\)
i.\(\frac{3}{2x+6}-\frac{x-6}{2x^2+6x}\)
giải hệ pt :
\(\hept{\begin{cases}3x^2+6xy+9y^2+\left(x+2y\right)^2\sqrt{x+2y}-3\left(x+2y\right)\sqrt{x+2y}-4\left(x+2y\right)+4\sqrt{x+2y}=0\\\left(\frac{\sqrt[3]{x^2-y^2}}{\sqrt[4]{x}}+\sqrt[4]{\frac{x}{y}}\right)^{2017}+\left(\sqrt[3]{\frac{x}{y}}-\sqrt[4]{\frac{y}{x}}\right)^{2018}=1\end{cases}}\)