So sánh:
a)
\(\dfrac{x-y}{x+y} và \dfrac{x^2-y^2}{x^2+y^2}\) với x>y>0
b) 216 và (2+1)(22+1)(24+1)(28+1)
Giải chi tiết giúp mk vs!!!😃 Thanks nhìu ạ
giải pt
\(\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{5}{6}\\\dfrac{\dfrac{2}{3}}{x}+\dfrac{\dfrac{2}{3}}{y}+\dfrac{\dfrac{8}{9}}{y}=1\end{matrix}\right.\)
giúp mình giải chi tiết với nha đừng làm tắt ok thanks
\(\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{5}{6}\\\dfrac{\dfrac{2}{3}}{x}+\dfrac{\dfrac{2}{3}}{y}+\dfrac{\dfrac{8}{9}}{y}=1\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{5}{6}\\\dfrac{\dfrac{2}{3}}{x}+\dfrac{\dfrac{14}{9}}{y}=1\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{5}{6}\left(1\right)\\\dfrac{2}{3x}+\dfrac{14}{9y}=1\left(2\right)\end{matrix}\right.\)
Nhân cả hai vế (1) cho \(\dfrac{2}{3}\) ta có: \(\left\{{}\begin{matrix}\dfrac{2}{3x}+\dfrac{2}{3y}=\dfrac{5.2}{6.3}\\\dfrac{2}{3x}+\dfrac{14}{9y}=1\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2}{3x}+\dfrac{2}{3y}=\dfrac{10}{18}\left(3\right)\\\dfrac{2}{3x}+\dfrac{14}{9y}=1\left(4\right)\end{matrix}\right.\)
Lấy (4) trừ (3) ta có:
\(\dfrac{14}{9y}-\dfrac{2}{3y}=1-\dfrac{10}{18}\)\(\Leftrightarrow\dfrac{8}{9y}=\dfrac{4}{9}\)\(\Leftrightarrow y=2\Rightarrow x=\dfrac{1}{\dfrac{5}{6}-\dfrac{1}{2}}=3\)
Cho A = \(\dfrac{x+y-2\sqrt{xy}}{x-y}\left(x\ge0;y\ge0;x\ne y\right)\)
1) Chứng minh A = \(\dfrac{\sqrt{x}-\sqrt{y}}{\sqrt{x}+\sqrt{y}}\)
2) Tính A với x = \(3+2\sqrt{2}\) và y = \(3-2\sqrt{2}\)
LÀM CHI TIẾT GIÚP MK NHÉ!
1: \(A=\dfrac{x-2\sqrt{xy}+y}{x-y}=\dfrac{\left(\sqrt{x}-\sqrt{y}\right)^2}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}=\dfrac{\sqrt{x}-\sqrt{y}}{\sqrt{x}+\sqrt{y}}\)
2: Thay \(x=3+2\sqrt{2}\) và \(y=3-2\sqrt{2}\) vào A, ta được:
\(A=\dfrac{\sqrt{2}+1-\sqrt{2}+1}{\sqrt{2}+1+\sqrt{2}-1}=\dfrac{2}{2\sqrt{2}}=\dfrac{\sqrt{2}}{2}\)
Tìm m để đồ thị của hàm số \(y=x-m\) đi qua điểm \(M_{\left(\dfrac{1}{3};\dfrac{1}{2}\right)}\)
A. \(\dfrac{1}{6}\)
B. \(-\dfrac{1}{6}\)
C. \(\dfrac{1}{3}\)
D. \(\dfrac{1}{2}\)
Cần cách giải chi tiết và đầy đủ ạ
giúp mk với tối nay phải nộp rồi
Chọn B. Thay \(\dfrac{1}{3}\)vào x và \(\dfrac{1}{2}\)vào y
giải để ra được m
(1) giải hpt:
a) \(\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{6}\\\dfrac{8}{x}+\dfrac{5}{y}=1\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}\dfrac{x-1}{2}-y=1\\2x+y=1\end{matrix}\right.\)
giúp mk vs ạ
a.\(\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{6}\\\dfrac{8}{x}+\dfrac{5}{y}=1\end{matrix}\right.\)
\(ĐK:x;y\ne0\)
Đặt \(\left\{{}\begin{matrix}\dfrac{1}{x}=a\\\dfrac{1}{y}=b\end{matrix}\right.\)
hpt trở thành:
\(\left\{{}\begin{matrix}a+b=\dfrac{1}{6}\\8a+5b=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{1}{18}\\b=\dfrac{1}{9}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{x}=\dfrac{1}{18}\\\dfrac{1}{y}=\dfrac{1}{9}\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=18\\y=9\end{matrix}\right.\) ( tm )
Vậy nghiệm hpt: \(\left\{{}\begin{matrix}x=18\\y=9\end{matrix}\right.\)
b.\(\left\{{}\begin{matrix}\dfrac{x-1}{2}-y=1\\2x+y=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x-1}{2}+2x=2\\2x+y=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x-1+4x}{2}=\dfrac{4}{2}\\2x+y=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}5x=5\\2x+y=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=1\\2.1+y=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\)
a.\(\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{6}\\\dfrac{8}{x}+\dfrac{5}{y}=1\end{matrix}\right.\)
\(ĐK:x;y\ne0\)
Đặt \(\left\{{}\begin{matrix}\dfrac{1}{x}=a\\\dfrac{1}{y}=b\end{matrix}\right.\)
hpt trở thành:
\(\left\{{}\begin{matrix}a+b=\dfrac{1}{6}\\8a+5b=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{1}{18}\\b=\dfrac{1}{9}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{x}=\dfrac{1}{18}\\\dfrac{1}{y}=\dfrac{1}{9}\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=18\\y=9\end{matrix}\right.\) ( tm )
Vậy nghiệm hpt: \(\left\{{}\begin{matrix}x=18\\y=9\end{matrix}\right.\)
b.\(\left\{{}\begin{matrix}\dfrac{x-1}{2}-y=1\\2x+y=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x-1}{2}+2x=2\\2x+y=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x-1+4x}{2}=\dfrac{4}{2}\\2x+y=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}5x=5\\2x+y=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=1\\2.1+y=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\)
Chứng minh rằng: A = \(\dfrac{\left(\sqrt{x}-\sqrt{y}\right)^2+4\sqrt{xy}}{\sqrt{x}+\sqrt{y}}-\dfrac{x\sqrt{y}-y\sqrt{x}}{\sqrt{xy}}\)không phụ thuộc vào x;y với x > 0 và y > 0
Các bạn lm chi tiết giúp mk nhé!
\(A=\dfrac{x-2\sqrt{xy}+y+4\sqrt{xy}}{\sqrt{x}+\sqrt{y}}-\dfrac{\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{xy}}\\ A=\dfrac{\left(\sqrt{x}+\sqrt{y}\right)^2}{\sqrt{x}+\sqrt{y}}-\sqrt{x}+\sqrt{y}\\ A=\sqrt{x}+\sqrt{y}-\sqrt{x}+\sqrt{y}=2\sqrt{y}\)
Đề sai
\(A=\dfrac{\left(\sqrt{x}-\sqrt{y}\right)^2+4\sqrt{xy}}{\sqrt{x}+\sqrt{y}}+\dfrac{x\sqrt{y}-y\sqrt{x}}{\sqrt{xy}}\)
\(=\sqrt{x}+\sqrt{y}+\sqrt{x}-\sqrt{y}\)
\(=2\sqrt{x}\)
Cho x, y, z đôi một khác nhau và \(\dfrac{1}{x}\)+\(\dfrac{1}{y}\)+\(\dfrac{1}{z}\) = 0
Tính giá trị của biểu thức: M = \(\dfrac{yz}{x^2+2yz}+\dfrac{xz}{y^2+2xz}+\dfrac{xy}{z^2+2xy}\)
Giúp mk giải bài này với, khó quá :((
Bài này ez thôi, làm mãi rồi.
Theo đề bài, ta có: \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\)
=>\(\dfrac{xy+yz+xz}{xyz}=0\)
=> xy+yz+zx=0
=> \(\left\{{}\begin{matrix}xy=-yz-zx\\yz=-xy-zx\\zx=-xy-yz\end{matrix}\right.\)
Ta có: x2+2yz=x2+yz-xy-zx=(x-y)(x-z)
y2+2xz=y2+xz-xy-yz=(x-y)(z-y)
z2+2xy=z2+xy-yz-xz=(x-z)(y-z)
=> \(\dfrac{yz}{\left(x-y\right)\left(x-z\right)}+\dfrac{xz}{\left(x-y\right)\left(z-y\right)}+\dfrac{xy}{\left(x-z\right)\left(y-z\right)}=\dfrac{yz\left(y-z\right)-xz\left(x-z\right)+xy\left(x-y\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}=\dfrac{\left(x-y\right)\left(x-z\right)\left(y-z\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}=1\)
Giải hệ pt:
a)(x+√(x^2+4))(y+√(y^2+1))=2 và 27x^6=x^3-8y+2
b)(8x-3)√(2x-1) -y-4y^3=0 và 4x^2-8x+2y^3+y^2-2y+3=0
c) x(1+y-x)=-2y^2-y và x(√2y -2)=y(√(x-1)-2)
d) √(x+2y)+√(2x-y)+x^2y=√x+√3y+xy^2 và 2(1-y)√(x^2+2y-1)=y^2-2x-1
e)(y-2x+√y-√x)/√xy +1=0 và √(1-xy) +x^2-y^2=0
CÁC BẠN ƠI..GIÚP MK VS Ạ...MAI MK HOK R...CẢM ƠM TRƯỚC Ạ...☺️☺️☺️
So sánh 2 biểu thức sau A=\(\dfrac{y^2\left(x+1\right)+\left(x+1\right)}{y^2+1}\) và B=\(\dfrac{y^2\left(x-1\right)+2x-x}{y^2+2}\)
Ta có :
A = \(\dfrac{\text{y^2 ( x + 1 ) + ( x + 1 ) }}{y^2+1}\) = \(\dfrac{\left(y^2+1\right)\left(x+1\right)}{y^2+1}\) = x+1 (1)
B = \(\dfrac{y^2\left(x-1\right)+2x-x}{y^2+2}=\dfrac{\left(y^2+2\right)\left(x-1\right)}{y^2+2}=x-1\) (2)
Từ (1) và (2)
=> A > B
\(\dfrac{\text{y^2 ( x + 1 ) + ( x + 1 ) }}{y^2+1}\) = \(\dfrac{\left(y^2+1\right)\left(x+1\right)}{y^2+1}\)
1, Cho x; y; z ≠0 và \(\dfrac{1}{x}\) + \(\dfrac{1}{y}\)+ \(\dfrac{1}{z}\)=\(\dfrac{2}{2x+y+2z}\). Cmr: (2x+y)(y+2z)(z+x)= 0
2, Cho \(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}=1\). Cmr: \(\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}=0\)
Gấp ạ, ai giúp mình với!!!!
2: Ta có: \(\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}=\dfrac{a\left(a+b+c\right)}{b+c}+\dfrac{b\left(a+b+c\right)}{c+a}+\dfrac{c\left(a+b+c\right)}{a+b}-a-b-c=\left(a+b+c\right)\left(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\right)=a+b+c-a-b-c=0\)
1: Sửa đề: Cho \(x,y,z\ne0\) và \(\dfrac{1}{x}+\dfrac{2}{y}+\dfrac{1}{z}=\dfrac{2}{2x+y+2z}\).
CM:....
Đặt 2x = x', 2z = z'.
Ta có: \(\dfrac{2}{x'}+\dfrac{2}{y}+\dfrac{2}{z'}=\dfrac{2}{x'+y+z'}\)
\(\Leftrightarrow\dfrac{1}{x'}+\dfrac{1}{y}+\dfrac{1}{z'}=\dfrac{1}{x'+y+z'}\)
\(\Leftrightarrow\dfrac{1}{x'}-\dfrac{1}{x'+y+z'}+\dfrac{1}{y}+\dfrac{1}{z'}=0\)
\(\Leftrightarrow\dfrac{y+z'}{x'\left(x'+y+z'\right)}+\dfrac{y+z'}{yz'}=0\)
\(\Leftrightarrow\dfrac{\left(y+z'\right)\left(yz'+x'^2+x'y+x'z'\right)}{x'yz'\left(x'+y+z'\right)}=0\)
\(\Leftrightarrow\dfrac{\left(x'+y\right)\left(y+z'\right)\left(z'+x'\right)}{x'yz'\left(x'+y+z'\right)}=0\Leftrightarrow\left(2x+y\right)\left(y+2z\right)\left(2z+2x\right)=0\Leftrightarrow\left(2x+y\right)\left(y+2z\right)\left(z+x\right)=0\left(đpcm\right)\)