\(\frac{\text{1}}{a^2-5a+6}\)+ \(\frac{\text{1}}{a^2-7a+12}\)+ \(\frac{1}{\text{}\text{a}\text{ }^2-9a+20}\)+ \(\frac{1}{^2-11a+30}\)
Rút gọn biểu thức
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
Rút gọn biểu thức
A=\(\frac{1}{a^2-5a+6}+\frac{1}{a^2-7a+12}+\frac{1}{a^2-9a+20}........\)
Rút gọn: \(\frac{1}{a^2-5a+6}+\frac{1}{a^2-7a+12}+\frac{1}{a^2-9a+20}+\frac{1}{a^2-11a+30}\)
Đăỵ tổng là A
\(\Rightarrow A=\frac{1}{a^2-5a-4+10}+\frac{1}{a^2-7a-16+28}+\frac{1}{a^2-9a-25+45}+\frac{1}{a^2-11a-36+66}\)
\(\Rightarrow A=\frac{1}{\left(a^2-4\right)-\left(5a-10\right)}+\frac{1}{\left(a^2-16\right)-\left(7a-28\right)}+\frac{1}{\left(a^2-25\right)-\left(9a-45\right)}+\frac{1}{\left(a^2-36\right)-\left(11a-66\right)}\)
\(\Rightarrow A=\frac{1}{\left(a+2\right)\left(a-2\right)-5\left(a-2\right)}+\frac{1}{\left(a+4\right)\left(a-4\right)-7\left(a-4\right)}+\frac{1}{\left(a-5\right)\left(a+5\right)-9\left(a-5\right)}+\frac{1}{\left(a-6\right)\left(a+6\right)-11\left(a-6\right)}\)
\(\Rightarrow A=\frac{1}{\left(a-2\right)\left(a-3\right)}+\frac{1}{\left(a-4\right)\left(a-3\right)}+\frac{1}{\left(a-5\right)\left(a-4\right)}+\frac{1}{\left(a-6\right)\left(a-5\right)}\)
\(\Rightarrow A=\frac{1}{a-3}-\frac{1}{a-2}+\frac{1}{a-4}-\frac{1}{a-3}+\frac{1}{a-5}-\frac{1}{a-4}+\frac{1}{a-6}-\frac{1}{a-5}\)
\(\Rightarrow A=\frac{1}{a-6}-\frac{1}{a-2}\)
\(\Rightarrow A=\frac{\left(a-2\right)-\left(a-6\right)}{\left(a-6\right)\left(a-2\right)}=\frac{4}{\left(a-6\right)\left(a-2\right)}\)
Rút gọn biểu thức:
A= \(\frac{1}{a^2-5a+6}+\frac{1}{a^2-7a+12}+\frac{1}{a^2-9a+20}\)
\(\frac{3}{\left(a-2\right)\left(a-3\right)}\). minh khong chac dau nha. neu sai thi thoi.
theo đề ta có :
A= (1 /a2 - 2a -3a + 6) + ( 1 /a2 -3a -4a + 12 ) + ( 1 /a2 - 4a -5a + 20 )
\(\Leftrightarrow\) A = (1 /( a -2 )(a-3 ))+ (1/(a-3)(a-4))+(1/(a-4)(a-5)).
\(\Leftrightarrow\)A = (1/a-2) - (1/ a-3) +(1/ a-3) -(1/ a-4) +(1/ a-4) - (1/ a-5)
\(\Leftrightarrow\)A = (1/a-2) - (1/ a-5)
\(\Leftrightarrow\)A = a-5-a+2 / (a-2)(a-5)
\(\Leftrightarrow\)A= 3 / ( a -2 )(a-5)
chúc bạn học tốt
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
Cho biểu thức
A= \(\text{[}1-\frac{\sqrt{x}}{1+\sqrt{x}}\text{]}:\text{[}\frac{\sqrt{x}+3}{\sqrt{x}+2}+\frac{\sqrt{x}+2}{3-\sqrt{x}}+\frac{\sqrt{x}+2}{x-5\sqrt{x}+6}\)
a, Rút gọn A
b, Tìm x để A<0
a: \(A=\dfrac{1}{\sqrt{x}+1}:\left(\dfrac{x-9-x+4+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\right)\)
\(=\dfrac{1}{\sqrt{x}+1}\cdot\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}{\sqrt{x}-3}\)
\(=\dfrac{\sqrt{x}-2}{\sqrt{x}+1}\)
b: Để A<0 thì \(\sqrt{x}-2< 0\)
hay 0<x<4
Cho biểu thức
A=\(\text{[}1-\frac{\sqrt{x}}{1+\sqrt{x}}\text{]}:\text{[}\frac{\sqrt{x}+3}{\sqrt{x}-2}+\frac{\sqrt{x}+2}{3-\sqrt{x}}+\frac{\sqrt{x}+2}{x-5\sqrt{x}+6}\)
a, Rút gọn A
b, Tìm x để A= \(\frac{1}{2}\)
a) A= (\(\left(\frac{1+\sqrt{x}}{1+\sqrt{x}}-\frac{\sqrt{x}}{1+\sqrt{x}}\right):\left(\frac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x-2}\right)}+\frac{\sqrt{x}+2}{x-2\sqrt{x}-3\sqrt{x}+6}\right)\)
A=\(\left(\frac{1+\sqrt{x}-\sqrt{x}}{1+\sqrt{x}}\right):\left(\frac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}+\frac{\sqrt{x}+2}{\sqrt{x}\left(\sqrt{x}-2\right)-3\left(\sqrt{x}-2\right)}\right)\)
A= \(\left(\frac{1}{1+\sqrt{x}}\right):\left(\frac{x-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\frac{x-4}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}+\frac{\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\right)\)
A=\(\left(\frac{1}{1+\sqrt{x}}\right):\left(\frac{x-9-x+4+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\right)\)
A=\(\left(\frac{1}{1+\sqrt{x}}\right):\left(\frac{\sqrt{x}-3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\right)\)
A=\(\frac{\sqrt{x}-2}{\sqrt{x}+1}\)
b) Để A = \(\frac{1}{2}\)
thì \(\frac{\sqrt{x}-2}{\sqrt{x}+1}=\frac{1}{2}\)
=> 2\(\sqrt{x}-4\)=\(\sqrt{x}+1\)
=> \(\sqrt{x}=5\)
=> x = 25
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
\(A=\left(\frac{x+1}{x-1}-\frac{x-1}{x+1}+\frac{x^2-4x-1}{x^2-1}\right)\div\frac{x}{x+2019}\)
ĐK : x ≠ ±1 ; x ≠ 0 ; x ≠ -2019
\(=\left(\frac{\left(x+1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}-\frac{\left(x-1\right)\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}+\frac{x^2-4x-1}{\left(x-1\right)\left(x+1\right)}\right)\times\frac{x+2019}{x}\)
\(=\left(\frac{x^2+2x+1}{\left(x-1\right)\left(x+1\right)}-\frac{x^2-2x+1}{\left(x-1\right)\left(x+1\right)}+\frac{x^2-4x-1}{\left(x-1\right)\left(x+1\right)}\right)\times\frac{x+2019}{x}\)
\(=\left(\frac{x^2+2x+1-x^2+2x-1+x^2-4x-1}{\left(x-1\right)\left(x+1\right)}\right)\times\frac{x+2019}{x}\)
\(=\frac{x^2-1}{x^2-1}\times\frac{x+2019}{x}=\frac{x+2019}{x}\)
b. \(A=\frac{x+2019}{x}=1+\frac{2019}{x}\) đạt giá trị lớn nhất
<=> \(\frac{2019}{x}\) đạt giá trị lớn nhất
<=> \(\hept{\begin{cases}x>0\\x\in Z\end{cases}}\) và x đạt giá trị bé nhất
<=> x = 1
Khi đó A = 2020
A=1- (\(\text{ }\frac{\text{2x^2 - 1+x}}{\text{1-x^2}}\text{+}\text{ }\frac{\text{2x^3 - x +x^2}}{\text{1+x^2}}\)) * \(\frac{\text{(((1-x)(x^2-x)}}{\text{2x - 1}}\)
Rút gọn A và Cm A < 4/3