Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
1) Vì theo đề bài \(\frac{x-2}{x-6}>0\Rightarrow x\ne0\)
Gọi phân số là \(\frac{a}{b}\)với \(a>b\) (vì tử số lớn hơn mẫu số thì phân số sẽ lớn hơn 1)
\(\Rightarrow x\ge6\)
2) Ta có: \(\frac{3x+9}{x-4}\) có giá trị nguyên . Với 3x + 9 > x - 4
Nếu x = 1 thì \(\frac{3x+9}{x-4}=\frac{31+9}{1-4}=\frac{40}{-31,3333}\) (loại)
Nếu x = 2 thì \(\frac{3x+9}{x-4}=\frac{32+9}{2-4}=\frac{41}{-2}=-20,5\) (loại)
Nếu x = 3 thì \(\frac{3x+9}{x-4}=\frac{33+9}{3-4}=\frac{42}{-1}=-42\)(chọn)
Nếu x = 4 thì \(\frac{3x+9}{x-4}=\frac{34+9}{4-4}=\frac{43}{0}\)(chọn)
Nếu x = 5 thì \(\frac{3x+9}{x-4}=\frac{35+9}{5-4}=\frac{44}{1}=44\)chọn
..và còn nhiều giá trị khác nữa...
Suy ra x = {-3 ; -4 ; -5 ; 3 ; 4 ; 5 ...}Tương tự ta có bảng sau:
| x nguyên dương | 3 | 4 | 5 |
| x nguyên âm | -3 | -4 | -5 |
Bài 3. Bí rồi, mình mới lớp 6 thôi!
bài 3: đạt B=\(\frac{1}{2}:\left(-1\frac{1}{2}\right):1\frac{1}{3}:\left(-1\frac{1}{4}\right):1\frac{1}{5}:\left(-1\frac{1}{6}\right)\):...:\(\left(-1\frac{1}{100}\right)\)
=\(\frac{1}{2}:\frac{-3}{2}:\frac{4}{3}:\frac{-5}{4}:\frac{6}{5}:\frac{-7}{6}:...:\frac{-101}{100}\)=\(\frac{1}{2}.\frac{-2}{3}.\frac{3}{4}.\frac{-4}{5}.\frac{5}{6}\frac{-6}{7}...\frac{-100}{101}\)(có 50 thừa số âm)
=\(\frac{1.2.3.4...100}{2.3.4...101}=\frac{1}{101}\)
vậy B=\(\frac{1}{101}\)
#HỌC TỐT#
1.
a) 13\(\frac{1}{3}\) : 1\(\frac{1}{3}\) = 26 : (2x - 1)
<=> \(\frac{40}{3}:\frac{4}{3}\) = 13x - 26
<=> 10 + 26 = 13x
<=> 13x = 36
<=> x = \(\frac{36}{13}\)
b) 0,2 : 1\(\frac{1}{5}\) = \(\frac{2}{3}\) : (6x + 7)
<=> \(\frac{1}{5}:\frac{6}{5}\) = \(\frac{1}{9}x\) : \(\frac{2}{21}\)
<=> \(\frac{1}{6}\) = \(\frac{1}{9}x\) : \(\frac{2}{21}\)
<=> \(\frac{1}{9}x\) = \(\frac{2}{21}.\frac{1}{6}\) = \(\frac{1}{63}\)
<=> x = \(\frac{1}{7}\)
c) \(\frac{37-x}{x+13}\) = \(\frac{3}{7}\)
<=> (37 - x) . 7 = 3.(x + 13)
<=> 119 - 7x = 3x + 39
<=> -7x - 3x = 39 - 119
<=> -10x = -80
<=> x = 8
d) \(\frac{x-1}{x+5}=\frac{6}{7}\)
<=> 7(x - 1) = 6(x + 5)
<=> 7x - 7 = 6x + 30
<=> 7x - 6x = 30 + 7
<=> x = 37
e)
2\(\frac{2}{\frac{3}{0,002}}\) = \(\frac{1\frac{1}{9}}{x}\)
<=> \(\frac{1501}{750}\) = \(\frac{10}{9}:x\)
<=> x = \(\frac{10}{9}:\frac{1501}{750}\) = \(\frac{2500}{4503}\)
Bài 2. đề sai
Bài 3.
a) 6,88 : x = \(\frac{12}{27}\)
<=> x = 6,88 : \(\frac{12}{27}\)
<=> x = 15,48
b) 8\(\frac{1}{3}\) : \(11\frac{2}{3}\) = 13 : 2x
<=> \(\frac{25}{3}:\frac{35}{3}\) = 13 : 2x
<=> \(\frac{5}{7}=13:2x\)
<=> 2x = \(13:\frac{5}{7}\) = \(\frac{91}{5}\)
<=> x = 9,1
ĐKXĐ \(x\ne0;x\ne1;x\ne-1\)
\(A=\frac{\left(x+1+1-x\right)}{\left(1-x^2\right)-\frac{5-x}{1-x^2}}:\frac{\left(1-2x\right)}{x^2-1}\)
\(A=\frac{\left(x-3\right)}{\left(1-x^2\right)}:\frac{\left(1-2x\right)}{\left(x^2-1\right)}\)
\(A=\frac{\left(3-x\right)}{\left(x^2-1\right)}:\frac{\left(1-2x\right)}{\left(x^2-1\right)}\)
\(A=\frac{\left(3x-2\right)}{1-2x}\)
\(a,ĐKXĐ:x\ne\pm1;x\ne\frac{1}{2}\)
\(A=\left(\frac{1}{x-1}+\frac{2}{x+1}-\frac{5-x}{1-x^{^2}}\right):\frac{1-2x}{x^2-1}\)
\(=\left(\frac{1}{x-1}+\frac{2}{x+1}+\frac{5-x}{\left(x-1\right)\left(x+1\right)}\right):\frac{1-2x}{\left(x-1\right)\left(x+1\right)}\)
\(=\frac{x+1+2\left(x-1\right)+5-x}{\left(x-1\right)\left(x+1\right)}:\frac{1-2x}{\left(x-1\right)\left(x+1\right)}\)
\(=\frac{2x+4}{\left(x-1\right)\left(x+1\right)}.\frac{\left(x-1\right)\left(x+1\right)}{1-2x}\)
\(=\frac{2x+4}{1-2x}\)
\(b,Vớix\ne\pm1;x\ne\frac{1}{2}\)ta có \(A=\frac{2x+4}{1-2x}=\frac{-1\left(1-2x\right)+5}{1-2x}=-1+\frac{5}{1-2x}\)
Với x thuộc Z để A nguyên thì \(5⋮1-2x\Rightarrow1-2x\inƯ\left\{5\right\}=\left\{\pm1;\pm5\right\}\)
Với 1-2x=1 => x= 0(TMĐKXĐ)
với 1-2x=-1 => x=1(loại)
với 1-2x=5 => x=-2(tmđkxđ)
với 1-2x=-5 => x=3(tmđkxđ)
Vậy với \(x\in\left\{0;-2;-3\right\}\)thì A nguyên
a) ĐKXĐ: \(x\ne-1;0;1.\)Ta có:
\(A=\left[\frac{2}{\left(x+1\right)^3}\left(\frac{1}{x}+1\right)+\frac{1}{x^2+2x+1}\left(\frac{1}{x^2}+1\right)\right]:\frac{x-1}{x^3}\)
\(=\left[\frac{2}{\left(x+1\right)^3}\cdot\frac{x+1}{x}+\frac{1}{\left(x+1\right)^2}\cdot\frac{x^2+1}{x^2}\right]\cdot\frac{x^3}{x-1}\)
\(=\left[\frac{2}{x\left(x+1\right)^2}+\frac{x^2+1}{x^2\left(x+1\right)^2}\right]\cdot\frac{x^3}{x-1}\)
\(=\left[\frac{2x}{x^2\left(x+1\right)^2}+\frac{x^2+1}{x^2\left(x+1\right)^2}\right]\cdot\frac{x^3}{x-1}\)
\(=\frac{2x+x^2+1}{x^2\left(x+1\right)^2}\cdot\frac{x^3}{x-1}\)
\(=\frac{\left(x+1\right)^2\cdot x}{\left(x+1\right)^2\left(x-1\right)}=\frac{x}{x-1}.\)
Vậy \(A=\frac{x}{x-1}\)với \(x\ne-1;0;1.\)
b) A < 1 \(\Leftrightarrow\frac{x}{x-1}< 1\Leftrightarrow\frac{x}{x-1}-1< 0\Leftrightarrow\frac{x}{x-1}-\frac{x-1}{x-1}< 0\)\(\Leftrightarrow\frac{1}{x-1}< 0\)
\(\Leftrightarrow x-1< 0\)(do 1 > 0)\(\Leftrightarrow x< 1.\)
Kết hợp ĐKXĐ, A < 1 khi \(x< 1\)và \(x\ne-1;0.\)
c) \(A\inℤ\Leftrightarrow\frac{x}{x-1}\inℤ.\)Mà \(x\inℤ\)\(\Rightarrow x⋮\left(x-1\right)\Rightarrow\left(x-1+1\right)⋮\left(x-1\right)\Rightarrow1⋮\left(x-1\right)\Rightarrow\left(x-1\right)\inƯ\left(1\right)=\left\{1;-1\right\}.\)Ta lập bảng sau:
| \(x-1\) | 1 | -1 |
| \(x\) | 2 | 0 |
| Kết luận | x thoả mãn ĐKXĐ | x không thoả mãn ĐKXĐ |
Vậy để A nguyên thì x = 2.
Ta có \(A=[\frac{2}{\left(x+1\right)^3}\left(\frac{1}{x}+1\right)+\frac{1}{x^2+2x+1}\left(\frac{1}{x^2}+1\right)]:\frac{x-1}{x^3}\)
\(\Leftrightarrow A=\left[\frac{2}{\left(x+1\right)^3}.\frac{x+1}{x}+\frac{1}{\left(x+1\right)^2}.\frac{x^2+1}{x^2}\right].\frac{x^3}{x-1}\)
\(\Leftrightarrow A=\left[\frac{2x+x^2+1}{x^2\left(x+1\right)^2}\right].\frac{x^3}{x+1}=\frac{x}{x+1}\)
Để \(A=\frac{x}{x+1}< 1\Leftrightarrow\frac{1}{x+1}>0\Leftrightarrow x>-1\)
Để \(A=1-\frac{1}{x+1}\text{ nguyên thì }\frac{1}{x+1}\text{ nguyên hay }x\in\left\{-2,0\right\} \)
\(A=\left(\frac{x^2-16}{x-4}-1\right):\left(\frac{x-2}{x-3}+\frac{x+3}{x+1}+\frac{x+2-x^2}{x^2-2x-3}\right)\)ĐK : \(x\ne3;-1;4\)
\(=\left(\frac{\left(x-4\right)\left(x+4\right)}{x-4}-1\right):\left(\frac{\left(x-2\right)\left(x+1\right)}{\left(x-3\right)\left(x+1\right)}+\frac{\left(x+3\right)\left(x-3\right)}{\left(x-3\right)\left(x+1\right)}+\frac{x+2-x^2}{\left(x-3\right)\left(x+1\right)}\right)\)
\(=\left(x-3\right):\left(\frac{x^2-x-2+x^2-9+x+2-x^2}{\left(x-3\right)\left(x+1\right)}\right)=\left(x-3\right):\left(\frac{x^2-9}{\left(x-3\right)\left(x-1\right)}\right)\)thơm thế :))
\(=\left(x-3\right):\left(\frac{\left(x-3\right)\left(x+3\right)}{\left(x-3\right)\left(x-1\right)}\right)=\left(x-3\right).\frac{x-1}{x+3}=\frac{\left(x-3\right)\left(x-1\right)}{x+3}\)
1) đk: \(x\ne\left\{-1;3;4\right\}\)
Ta có:
\(A=\left(\frac{x^2-16}{x-4}-1\right)\div\left(\frac{x-2}{x-3}+\frac{x+3}{x+1}+\frac{x+2-x^2}{x^2-2x-3}\right)\)
\(A=\left[\frac{\left(x-4\right)\left(x+4\right)}{x-4}-1\right]\div\frac{\left(x-2\right)\left(x+1\right)+\left(x+3\right)\left(x-3\right)+x+2-x^2}{\left(x+1\right)\left(x-3\right)}\)
\(A=\left(x+4-1\right)\div\frac{x^2-x-2+x^2-9-x^2+x+2}{\left(x+1\right)\left(x-3\right)}\)
\(A=\left(x+3\right)\div\frac{x^2-9}{\left(x+1\right)\left(x-3\right)}\)
\(A=\left(x+3\right)\cdot\frac{\left(x+1\right)\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}\)
\(A=x+1\)
2) Ta có: \(\frac{A}{x^2+x+1}=\frac{x+1}{x^2+x+1}\)
Để \(\frac{A}{x^2+x+1}\) nguyên thì \(\left(x+1\right)⋮\left(x^2+x+1\right)\Leftrightarrow\left(x+1\right)^2⋮\left(x^2+x+1\right)\)
\(\Rightarrow\left(x+1\right)^2-\left(x^2+x+1\right)⋮\left(x^2+x+1\right)\)
\(\Rightarrow x⋮\left(x^2+x+1\right)\Rightarrow1⋮x^2+x+1\)
\(\Rightarrow x^2+x+1\in\left\{-1;1\right\}\Rightarrow x^2+x+1=1\Leftrightarrow x^2+x=0\Rightarrow\orbr{\begin{cases}x=-1\left(ktm\right)\\x=0\left(tm\right)\end{cases}}\)
Vậy x = 0
