\(\frac{\text{1}}{6}\)x + \(\frac{\text{1}}{\text{1}0}\) x - \(\frac{4}{5}\) x + 1 = 0
Những câu hỏi liên quan
Ai giải giúp mấy bài toán vsBài 1:Asqrt{frac{1}{text{√}2+1}-frac{text{√}8-text{√}10}{2-text{√}5}}Bfrac{5text{√}5}{text{√}5+2}+frac{text{√}5}{text{√}5-1}-frac{3text{√}5}{3+text{√}5}Bài 2 rút gọn biểu thứcAleft(frac{x+sqrt[]{xy}}{text{√}x+text{√}y}-2right):frac{1}{text{√}x+2} với x :y 0Bleft(frac{a}{a-2text{√}a}+frac{a}{text{√}a-2}right):frac{text{√}a+1}{a-4text{√}a+4}Bài 3 cho biểu thứcPleft(frac{x-2}{x+2text{√}x}+frac{1}{text{√}x+2}right)frac{text{√}x+1}{text{√}x-1}a)Rút gọn Pb)tìm x để Ptext{√}...
Đọc tiếp
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
\(x\text{(\frac{1}{5} +\frac{1}{4})-\text{( \frac{1}{7}-\frac{1}{8})=0}}\)
ở sao đăng câu hỏi mà ko có bài vậy bạn troll ak
bai 1 \(\frac{-3}{\text{2}}+\frac{5}{7}+\frac{-31}{14}< hoac=\text{x}< \frac{1}{\text{2}}+\frac{1}{3}+\frac{1}{6}\)\(\frac{1}{6}\)
bai 2 \(\frac{\text{x}+4}{\text{x}-\text{2}}+\frac{\text{2}\text{x}-5}{\text{x}-\text{2}}\)la so nguyen
Bài 1 mk ko hiểu đề cho lắm
Bài 2 :
Đặt \(A=\frac{x+4}{x-2}+\frac{2x-5}{x-2}\)
Ta có :
\(\frac{x+4}{x-2}+\frac{2x-5}{x-2}=\frac{x+4+2x-5}{x-2}=\frac{3x-1}{x-2}=\frac{3x-6+5}{x-2}=\frac{3\left(x-2\right)}{x-2}+\frac{5}{x-2}=3+\frac{5}{x-2}\)
Để \(A\) là số nguyên thì \(\frac{5}{x-2}\) phải là số nguyên \(\Rightarrow\) \(5⋮\left(x-2\right)\) \(\Rightarrow\) \(\left(x-2\right)\inƯ\left(5\right)\)
Mà \(Ư\left(5\right)=\left\{1;-1;5;-5\right\}\)
Do đó :
| \(x-2\) | \(1\) | \(-1\) | \(5\) | \(-5\) |
| \(x\) | \(3\) | \(1\) | \(7\) | \(-3\) |
Vậy \(x\in\left\{-3;1;3;7\right\}\) thì A là số nguyên
Chúc bạn học tốt ~
Đúng 0
Bình luận (0)
bai 1: Viết tập hợp A các số nguyên x biết:
cau hoi cua phung minh quan
Đúng 0
Bình luận (0)
Xem thêm câu trả lời
Thực hiện phép tínha) frac{text{x + 9}}{x^2 - 9}-frac{text{3}}{text{x^2 + 3x}}b) frac{text{3x + 5
}}{text{x^2 - 5x
}}+frac{text{ 25 - x
}}{text{25 - 5x
}}c) frac{text{3
}}{text{2x
}}+frac{text{3x - 3
}}{text{2x - 1
}}+frac{ 2x^2 + 1
}{text{4x^2 - 2x
}}d) frac{text{1}}{text{3x - 2 }}-frac{1}{text{3x + 2 }}- frac{text{3x - 6}}{text{4 - 9x^2}}e) frac{text{18 }}{text{(x - 3)(x^2 - 9) }}-frac{text{3 }}{text{x^2 - 6x + 9 }}-frac{text{x}}{text{x^2 - 9}}g) frac{text{x + 2 }}{text{x + 3 }}-frac{text{...
Đọc tiếp
Thực hiện phép tính
a) \(\frac{\text{x + 9}}{x^2 - 9}-\frac{\text{3}}{\text{x^2 + 3x}}\)
b) \(\frac{\text{3x + 5 }}{\text{x^2 - 5x }}+\frac{\text{ 25 - x }}{\text{25 - 5x }}\)
c) \(\frac{\text{3 }}{\text{2x }}+\frac{\text{3x - 3 }}{\text{2x - 1 }}+\frac{ 2x^2 + 1 }{\text{4x^2 - 2x }}\)
d) \(\frac{\text{1}}{\text{3x - 2 }}-\frac{1}{\text{3x + 2 }}- \frac{\text{3x - 6}}{\text{4 - 9x^2}}\)
e) \(\frac{\text{18 }}{\text{(x - 3)(x^2 - 9) }}-\frac{\text{3 }}{\text{x^2 - 6x + 9 }}-\frac{\text{x}}{\text{x^2 - 9}}\)
g) \(\frac{\text{x + 2 }}{\text{x + 3 }}-\frac{\text{5 }}{\text{x^2 + x - 6 }}+\frac{\text{1}}{\text{2 - x}}\)
h) \(\frac{\text{4x }}{\text{x + 2 }}-\frac{\text{3x }}{\text{x - 2 }}+\frac{\text{12x}}{\text{x^2 - 4}}\)
i) \(\frac{\text{ x + 1 }}{\text{ x - 1 }}-\frac{\text{ x - 1 }}{\text{ x + 1 }}-\frac{\text{4}}{\text{1 - x^2}}\)
k) \(\frac{\text{
3x + 21
}}{\text{
x^2 - 9
}}+\frac{\text{2 }}{\text{x + 3 }}-\frac{\text{3}}{\text{x - 3}}\)
Tính:1. frac{x^2}{x^2-x}-frac{x^2}{x+1}-frac{2text{x}}{x^2-1}2. frac{4x^2-3x+5}{x^3-1}-frac{1-2text{x}}{x^2+x+1}-frac{6}{x-1}3. frac{5}{2text{x}^2+6text{x}}-frac{4-3text{x}^2}{x^2-9}-34. frac{7}{8x^2-18}+frac{1}{2text{x}^2+3text{x}}-frac{1}{4text{x}-6}5. frac{1}{xleft(x+1right)}+frac{1}{left(x+1right)left(x+2right)}+...+frac{1}{left(x+9right)left(x+10right)}
Đọc tiếp
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. \(\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+...+\frac{1}{\left(x+9\right)\left(x+10\right)}\)
tính:
\(\frac{\text{x+1}}{10}+\frac{\text{x+1}}{11}-\frac{\text{x+1}}{13}+\frac{\text{x+1}1}{12}-\frac{\text{x+1}}{14}=0\)
B=\(\left(\frac{x\sqrt{x}}{x\text{+}\sqrt{x}\text{+}1}-\frac{1}{x\text{+}\sqrt{x}\text{+}1}\right):\frac{2}{\sqrt{x}\text{+}1}\)
Chứng minh A<0 với mọi 0<x<1
\(\frac{\text{1}}{5x8}\) + \(\frac{\text{1}}{8x\text{1}\text{1}}\)+ \(\frac{\text{1}}{\text{1}\text{1}x\text{1}4}\) + ..... + \(\frac{\text{1}}{x\left(x+3\right)}\)= \(\frac{\text{1}0\text{1}}{\text{1}540}\)
Ta có:
\(\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+...+\frac{1}{x\left(x+3\right)}=\frac{101}{1540}\)
\(\Rightarrow\frac{1}{3}.\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{x}-\frac{1}{x+3}\right)=\frac{101}{1540}\)
\(\Rightarrow\frac{1}{5}-\frac{1}{x+3}=\frac{101}{1540}.3=\frac{303}{1540}\)
\(\Rightarrow\frac{1}{x+3}=\frac{1}{5}-\frac{303}{1540}=\frac{1}{308}\)
\(\Rightarrow x+3=308\Leftrightarrow x=305\)
Đúng 0
Bình luận (0)
cho các số x,y,z khác 0 va thoả mãn :\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0.t\text{ính}gi\text{á}tr\text{ị}bi\text{ểu}th\text{ức}P=\frac{y+z}{x}+\frac{z+x}{y}+\frac{x+y}{z}\)
\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\)
\(\Leftrightarrow\hept{\begin{cases}\frac{1}{x}+\frac{1}{y}=-\frac{1}{z}\\\frac{1}{y}+\frac{1}{z}=-\frac{1}{x}\\\frac{1}{x}+\frac{1}{z}=-\frac{1}{y}\end{cases}}\)
\(P=\frac{y+z}{x}+\frac{z+x}{y}+\frac{x+y}{z}\)
\(=\frac{y}{x}+\frac{z}{x}+\frac{z}{y}+\frac{x}{y}+\frac{x}{z}+\frac{y}{z}\)
\(=y\left(\frac{1}{x}+\frac{1}{z}\right)+x\left(\frac{1}{z}+\frac{1}{y}\right)+z\left(\frac{1}{x}+\frac{1}{y}\right)\)
\(=y.\frac{-1}{y}+x.\frac{-1}{x}+z.\frac{-1}{z}\)
\(=-1-1-1=-3\)
Đúng 0
Bình luận (0)
P+3=\(\frac{y+z}{x}+1+\frac{x+z}{y}+1+\frac{x+y}{z}+1=\frac{x+y+z}{x}+\frac{x+y+z}{y}+\frac{x+y+z}{x}\)
P+3=\(\left(x+y+z\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=0.\left(x+y+z\right)=0\)
=> P=\(-3\)
Chuc ban hoc tot
Đúng 0
Bình luận (0)
Ta có : \(P=\frac{y+z}{x}+\frac{z+x}{y}+\frac{x+y}{z}\)
\(\Rightarrow P+3=\frac{y+z}{x}+1+\frac{z+x}{y}+1+\frac{x+y}{z}+1\)
\(\Rightarrow P+3=\frac{x+y+z}{x}+\frac{x+y+z}{y}+\frac{x+y+z}{z}\)
\(\Rightarrow P+3=\left(x+y+z\right).\frac{1}{x}+\left(x+y+z\right).\frac{1}{y}+\left(x+y+z\right).\frac{1}{z}\)
\(\Rightarrow P+3=\left(x+y+z\right).\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\)
\(\Rightarrow P+3=\left(x+y+z\right).0\)
\(\Rightarrow P+3=0\)
\(\Rightarrow P=-3\)
Vậy P = - 3
Đúng 0
Bình luận (0)