cho biểu thức \(A=\left(\frac{\text{x+1}}{\text{x-1}}-\frac{\text{x-1}}{\text{x+1}}+\frac{\text{x^2-4x-1}}{\text{x^2-1}}\right)\div\frac{x}{\text{x+2019}}\)
A) tìm đkxd và rút gọn
B) với \(x\in Z\), tìm GTLN của A
giúp mik câu b gấp nhé các bn
cho biểu thức\(A=\left(\frac{\text{x+1}}{\text{x-1}}-\frac{\text{x-1}}{\text{x+1}}+\frac{\text{x^2-4x-1)}}{\text{x^2-1}}\right)\div\frac{x}{x+2019}\)
a) tìm điều kiện xát định và rút gọn
b) với \(x\in z\)tìm GTLN của biểu thức
mik cần gấp câu b
Ai giải giúp mấy bài toán vs
Bài 1:
A=\(\sqrt{\frac{1}{\text{√}2+1}-\frac{\text{√}8-\text{√}10}{2-\text{√}5}}\)
B=\(\frac{5\text{√}5}{\text{√}5+2}+\frac{\text{√}5}{\text{√}5-1}-\frac{3\text{√}5}{3+\text{√}5}\)
Bài 2 rút gọn biểu thức
A=\(\left(\frac{x+\sqrt[]{xy}}{\text{√}x+\text{√}y}-2\right):\frac{1}{\text{√}x+2}\) với x :y >0
B=\(\left(\frac{a}{a-2\text{√}a}+\frac{a}{\text{√}a-2}\right):\frac{\text{√}a+1}{a-4\text{√}a+4}\)
Bài 3 cho biểu thức
P=\(\left(\frac{x-2}{x+2\text{√}x}+\frac{1}{\text{√}x+2}\right)\frac{\text{√}x+1}{\text{√}x-1}\)
a)Rút gọn P
b)tìm x để P=\(\text{√}x+\frac{5}{2}\)
bài 4 rút gọn biểu thức
A=\(\frac{1}{x+\text{√}x}+\frac{2\text{√}x}{x-1}-\frac{1}{x-\text{√}x}\)
B=\(\left(\frac{x}{x+3\text{√}x}+\frac{1}{\text{√}x+3}\right):\left(1-\frac{2}{\text{√}x}+\frac{6}{x+3\text{√}x}\right)\)
Bài 5
A=\(\left(\frac{2}{\text{√}x-3}-\frac{1}{\text{√}x+3}-\frac{x}{\text{√}x\left(x-9\right)}\right):\text{(√}x+3-\frac{x}{\text{√}x-3}\)
a)rút gọn A
b)tìm gtri x để A= -1/4
AI GIẢI GIÙM MÌNH ĐI MÌNH TẠ ƠN
Cho biểu thức
A=\(\orbr{ }\frac{x^2}{x^3-4x}+\frac{6}{6-3x}+\frac{1}{x+2}\text{] }\div\orbr{ }x-2+\frac{10-x^2}{x+2}\text{] }\)
a. Rút gọn A
b. Tìm giá trị của x để A có giá trị nguyên
\(A=\left[\frac{x^2}{x^3-4x}+\frac{6}{6-3x}+\frac{1}{x+2}\right]:\left[x-2+\frac{10-x^2}{x+2}\right]\) ĐKXĐ : \(x\ne0;x\ne\pm2\)
\(A=\left[\frac{x^2}{x\left(x+2\right)\left(x-2\right)}-\frac{6}{3\left(x-2\right)}+\frac{1}{x+2}\right]:\left[\frac{x^2-4}{x+2}+\frac{10-x^2}{x+2}\right]\)
\(A=\left[\frac{3x^2}{3x\left(x+2\right)\left(x-2\right)}-\frac{6x\left(x+2\right)}{3x\left(x+2\right)\left(x-2\right)}+\frac{3x\left(x+2\right)}{3x\left(x+2\right)\left(x-2\right)}\right]:\frac{6}{x+2}\)
\(A=\left[\frac{3x^2-6x^2-12x+3x^2+6x}{3x\left(x+2\right)\left(x-2\right)}\right].\frac{x+2}{6}\)
\(A=\frac{-x}{3x\left(x-2\right)}\)
\(A=\frac{-1}{3x-6}\)
\(\frac{1}{\text{x}^2+yz}+\frac{1}{y^2+\text{x}z}+\frac{1}{z^2+\text{x}y}\le\frac{1}{2}\left(\frac{1}{\text{x}y}+\frac{1}{yz}+\frac{1}{\text{x}z}\right)\)
đk: x;y;z dương nhé
áp dụng bđt cosi ta có:
\(x^2+yz>=2\sqrt{x^2yz}=2x\sqrt{yz};y^2+xz>=2\sqrt{y^2xz}=2y\sqrt{xz};z^2+xy=2\sqrt{z^2xy}=2z\sqrt{xy}\)
\(\Rightarrow\frac{1}{x^2+yz}< =\frac{1}{2x\sqrt{yz}};\frac{1}{y^2+xz}< =\frac{1}{2y\sqrt{xz}};\frac{1}{z^2+xy}< =\frac{1}{2z\sqrt{xy}}\)
\(\Rightarrow\frac{1}{x^2+yz}+\frac{1}{y^2+xz}+\frac{1}{z^2+xy}< =\frac{1}{2x\sqrt{yz}}+\frac{1}{2y\sqrt{xz}}+\frac{1}{2z\sqrt{xy}}=\frac{1}{2}\left(\frac{1}{x\sqrt{yz}}+\frac{1}{y\sqrt{xz}}+\frac{1}{z\sqrt{xy}}\right)\left(1\right)\)
áp dụng bđt cosi ta có:
\(\frac{1}{xy}+\frac{1}{xz}>=2\cdot\sqrt{\frac{1}{xy}\cdot\frac{1}{xz}}=\frac{2}{x\sqrt{yz}};\frac{1}{xy}+\frac{1}{yz}>=2\cdot\sqrt{\frac{1}{xy}\cdot\frac{1}{yz}}=\frac{2}{y\sqrt{xz}};\)
\(\frac{1}{yz}+\frac{1}{xz}>=2\cdot\sqrt{\frac{1}{yz}\cdot\frac{1}{xz}}=\frac{2}{z\sqrt{xy}}\)
\(\Rightarrow\frac{1}{xy}+\frac{1}{xz}+\frac{1}{xy}+\frac{1}{yz}+\frac{1}{yz}+\frac{1}{xz}=\frac{2}{xy}+\frac{2}{yz}+\frac{2}{xz}>=\frac{2}{x\sqrt{yz}}+\frac{2}{y\sqrt{xz}}+\frac{2}{z\sqrt{xy}}\)
\(\Rightarrow\frac{1}{xy}+\frac{1}{yz}+\frac{1}{xz}>=\frac{1}{x\sqrt{yz}}+\frac{1}{y\sqrt{xz}}+\frac{1}{z\sqrt{xy}}\)
\(\Rightarrow\frac{1}{2}\left(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{xz}\right)>=\frac{1}{2}\left(\frac{1}{x\sqrt{yz}}+\frac{1}{y\sqrt{xz}}+\frac{1}{z\sqrt{xy}}\right)\left(2\right)\)
từ \(\left(1\right);\left(2\right)\Rightarrow\frac{1}{x^2+yz}+\frac{1}{y^2+xz}+\frac{1}{z^2+xy}>=\frac{1}{2}\left(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{xz}\right)\left(đpcm\right)\)
dấu = xảy ra khi x=y=z
nhầm từ \(\left(1\right);\left(2\right)\Rightarrow\frac{1}{x^2+yz}+\frac{1}{y^2+xz}+\frac{1}{z^2+xy}< =\frac{1}{2}\left(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{xz}\right)\)
Cho biểu thức
A= \(\text{[}\frac{x+1}{x-1}-\frac{x-1}{x+1}\text{]}:\text{[}\frac{2}{x^2-1}-\frac{x}{x-1}+\frac{1}{x+1}\text{]}\)
a, Rút gọn A
b, Tính giá trị của A khi x =\(\sqrt{3+\sqrt{8}}\)
c, Tìm x để A= \(\sqrt{5}\)
ĐKXĐ : \(x\ne\pm1\)
a/ \(A=\left(\frac{x+1}{x-1}-\frac{x-1}{x+1}\right):\left(\frac{2}{x^2-1}-\frac{x}{x-1}+\frac{1}{x+1}\right)\)
\(=\frac{x^2+2x+1-\left(x^2-2x+1\right)}{\left(x-1\right)\left(x+1\right)}:\frac{2-x\left(x+1\right)+\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}\)
\(=\frac{4x}{\left(x-1\right)\left(x+1\right)}.\frac{\left(x-1\right)\left(x+1\right)}{1-x^2}=\frac{4x}{1-x^2}\)
b/ Ta có \(3+2\sqrt{2}=\left(\sqrt{2}+1\right)^2\Rightarrow\sqrt{3+\sqrt{8}}=\sqrt{2}+1\)
Suy ra : Nếu x = \(\sqrt{2}+1\) thì \(A=\frac{4\left(\sqrt{2}+1\right)}{1-\left(\sqrt{2}+1\right)^2}=\frac{4\left(\sqrt{2}+1\right)}{-\sqrt{2}.\sqrt{2}\left(\sqrt{2}+1\right)}=-\frac{4}{2}=-2\)
c/ \(A=\sqrt{5}\Rightarrow4x=\sqrt{5}\left(1-x^2\right)\Leftrightarrow\sqrt{5}x^2+4x-\sqrt{5}=0\)
Nhân cả hai vế của pt trên với \(\sqrt{5}\ne0\)
Được \(5x^2+4\sqrt{5}x-5=0\) . Đặt \(t=x\sqrt{5}\) pt trở thành \(t^2+4t-5=0\Leftrightarrow\left(t+5\right)\left(t-1\right)=0\) \(\Leftrightarrow\left[\begin{array}{nghiempt}t=1\\t=-5\end{array}\right.\)
Với t = 1 thì \(x=\frac{1}{\sqrt{5}}=\frac{\sqrt{5}}{5}\)
Với t = -5 thì \(x=-\frac{5}{\sqrt{5}}=-\sqrt{5}\)
\(A=\left[\frac{x^2+2x+1-x^2+2x-1}{x^2-1}\right]:\left[\frac{2-x^2-x+x-1}{x^2-1}\right]=\left[\frac{4x}{x^2-1}\right].\left[\frac{x^2-1}{1-x^2}\right]=\frac{4x}{1-x^2}\)
Tìm giá trị của biểu thức:
\(\frac{\left(x-2\right)\left(2\text{x}+2\text{x}^2\right)}{\left(x+1\right)\left(4\text{x}-x^3\right)}\)với x=\(\frac{-1}{2}\)
cho P =
\(\frac{x^2+2\text{x}}{2\text{x}+10}\)+ \(\frac{x-5}{x}\)- \(\frac{6\text{x}+5}{\left(2\text{x}+3\right)\left(2\text{x}-3\right)}\)
a, Rút gọn
b, tìm x để P=1;P=-3
1.Giải pt sau:(\(\sqrt{2}\) +2)(x\(\sqrt{2}\) -1)=2x\(\sqrt{2}\) -\(\sqrt{2}\)
2.Cho pt: 2(a-1).x-a(x-1)=2a+3
3.Giải pt sau:
a) \(\frac{2}{x+\frac{\text{1}}{\text{1}+\frac{x+\text{1}}{x-2}}}=\frac{6}{3x-\text{1}}\)
b) \(\frac{\frac{x+\text{1}}{x-\text{1}}-\frac{x-\text{1}}{x+\text{1}}}{\text{1}+\frac{x+\text{1}}{x-\text{1}}}=\frac{x-\text{1}}{2\left(x+\text{1}\right)}\)
1) Nhìn cái pt hết ham, nhưng bấm nghiệm đẹp v~`~
\(\left(\sqrt{2}+2\right)\left(x\sqrt{2}-1\right)=2x\sqrt{2}-\sqrt{2}\)
\(\Leftrightarrow\left(\sqrt{2}+2\right)\left(x\sqrt{2}-1\right)-2x\sqrt{2}+\sqrt{2}=0\)
\(\Leftrightarrow2x-\sqrt{2}+2x\sqrt{2}-2-2x\sqrt{2}+\sqrt{2}=0\)
\(\Leftrightarrow2x-2=0\Leftrightarrow2x=2\Rightarrow x=1\)
Mấy bài kia sao cái phương trình dài thê,s giải sao nổi
Tính:
1. \(\frac{x^2}{x^2-x}-\frac{x^2}{x+1}-\frac{2\text{x}}{x^2-1}\)
2. \(\frac{4x^2-3x+5}{x^3-1}-\frac{1-2\text{x}}{x^2+x+1}-\frac{6}{x-1}\)
3. \(\frac{5}{2\text{x}^2+6\text{x}}-\frac{4-3\text{x}^2}{x^2-9}-3\)
4. \(\frac{7}{8x^2-18}+\frac{1}{2\text{x}^2+3\text{x}}-\frac{1}{4\text{x}-6}\)
5. \(\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+...+\frac{1}{\left(x+9\right)\left(x+10\right)}\)