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a, \(\frac{3}{4}-x=\frac{1}{2}\Leftrightarrow x=\frac{3}{4}-\frac{1}{2}=\frac{1}{4}\)Vậy \(x=\frac{1}{4}\)
b, \(\left|x+\frac{2}{3}\right|=\frac{5}{6}\)
TH1 : \(x+\frac{2}{3}=\frac{5}{6}\Leftrightarrow x=\frac{5}{6}-\frac{2}{3}=\frac{1}{6}\)
TH2 : \(x+\frac{2}{3}=-\frac{5}{6}\Leftrightarrow x=-\frac{5}{6}-\frac{2}{3}=\frac{-9}{6}=\frac{-3}{2}\)
Vậy \(x=\left\{\frac{1}{6};-\frac{3}{2}\right\}\)
a,\(\frac{3}{4}-x=\frac{1}{2}\)
\(\Leftrightarrow x=\frac{3}{4}-\frac{1}{2}\)
\(\Leftrightarrow x=\frac{1}{4}\)
b,\(\left|x+\frac{2}{3}\right|=\frac{5}{6}\)
\(\Leftrightarrow x+\frac{2}{3}=\pm\frac{5}{6}\)
TH1:\(x+\frac{2}{3}=\frac{5}{6}\)
\(\Leftrightarrow x=\frac{5}{6}-\frac{2}{3}\)
\(\Leftrightarrow x=\frac{1}{6}\)
TH2:\(x+\frac{2}{3}=-\frac{5}{6}\)
\(\Leftrightarrow x=-\frac{5}{6}-\frac{2}{3}\)
\(\Leftrightarrow x=-\frac{3}{2}\)
a) \(\frac{x}{4}=\frac{16}{x^2}\)\(=>x^3=16.4\)\(=>x^3=64\)\(=>x=4\)
b) \(\frac{4}{3}:\frac{4}{5}=\frac{2}{3}.\left(\frac{1}{10}.x\right)\)\(=>\frac{4}{3}.\frac{5}{4}=\frac{2}{3}\left(\frac{1}{10}x\right)\)\(=>\frac{5}{3}=\frac{2}{3}\left(\frac{1}{10}x\right)\)\(=>\frac{5}{3}:\frac{2}{3}=\frac{1}{10}x\)\(=>\frac{5}{3}.\frac{3}{2}=\frac{1}{10}x\)\(=>\frac{5}{2}=\frac{1}{10}x\)\(=>x=\frac{5}{2}:\frac{1}{10}\)\(=>x=\frac{5}{2}.10\)\(=>x=25\)
vậy x=25
1.
a) \(\frac{x}{4}=\frac{16}{x^2}\)
\(\Rightarrow x^3=64\)
\(\Rightarrow x^3=4^3\)
\(\Rightarrow x=4\)
b) \(1\frac{1}{3}:0,8=\frac{2}{3}.\left(0,1.x\right)\)
\(\frac{5}{3}=\frac{2}{3}.\frac{x}{10}\)
\(\frac{x}{10}=\frac{5}{2}\)
\(\Rightarrow x=\frac{5.10}{2}=25\)
2.
\(A=\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}+\frac{1}{3^{99}}\)
\(3A=1+\frac{1}{3}+...+\frac{1}{3^{97}}+\frac{1}{3^{98}}\)
\(3A-A=\left(1+\frac{1}{3}+...+\frac{1}{3^{97}}+\frac{1}{3^{98}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}+\frac{1}{3^{99}}\right)\)
\(2A=1-\frac{1}{3^{99}}< 1\)
\(\Rightarrow A=\frac{1-\frac{1}{3^{99}}}{2}< \frac{1}{2}\)
\(\left(\frac{2}{3}x-1\right).\left(\frac{3}{4}x+\frac{1}{2}\right)=0\)
\(\Leftrightarrow\hept{\begin{cases}\frac{2}{3}x-1=0\\\frac{3}{4}x+\frac{1}{2}=0\end{cases}\Leftrightarrow\hept{\begin{cases}\frac{2}{3}x=1\\\frac{3}{4}x=\frac{-1}{2}\end{cases}\Leftrightarrow}\hept{\begin{cases}x=\frac{3}{2}\\x=-\frac{2}{3}\end{cases}}}\)
vậy \(x=\frac{3}{2}\)hoặc\(x=-\frac{2}{3}\)
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) \(\frac{3}{4}-\frac{-1}{2}+\frac{1}{3}=\frac{3}{4}+\frac{1}{2}+\frac{1}{3}=\frac{9}{12}+\frac{6}{12}+\frac{4}{12}=\frac{19}{12}\)
b) \(5\frac{5}{27}+\frac{7}{23}+\frac{1}{2}-\frac{5}{27}+\frac{16}{23}\)
\(=\frac{140}{27}-\frac{5}{27}+\frac{7}{23}+\frac{16}{23}+\frac{1}{2}\)
\(=\frac{135}{27}+\frac{23}{23}+\frac{1}{2}\)
\(=5+1+0,5=6,5\)
2) a) 1/2 + 2/3x = 1/4
=> 2/3x = 1/4 - 1/2
=> 2/3x = -1/4
=> x = -1/4 : 2/3
=> x = -3/8
b) 3/5 + 2/5 : x = 3 1/2
=> 3/5 + 2/5 : x = 7/2
=> 2/5 : x = 7/2 - 3/5
=> 2/5 : x = 29/10
=> x = 2/5 : 29/10
=> x = 4/29
c) x+4/2004 + x+3/2005 = x+2/2006 + x+1/2007
=> x+4/2004 + 1 + x+3/2005 + 1 = x+2/2006 + 1 + x+1/2007 + 1
=> x+2008/2004 + x+2008/2005 = x+2008/2006 + x+2008/2007
=> x+2008/2004 + x+2008/2005 - x+2008/2006 - x+2008/2007 = 0
=> (x+2008). (1/2004 + 1/2005 - 1/2006 - 1/2007) = 0
Vì 1/2004 + 1/2005 - 1/2006 - 1/2007 khác 0
Nên x + 2008 = 0 <=> x = -2008
Vậy x = -2008
1,a,\(\frac{3}{4}-\frac{-1}{2}+\frac{1}{3}=\frac{3}{4}+\frac{2}{4}+\frac{1}{3}=\frac{5}{4}+\frac{1}{3}=\frac{15}{12}+\frac{4}{12}=\frac{19}{12}\)
b, \(5\frac{5}{27}+\frac{7}{23}+\frac{1}{2}-\frac{5}{27}+\frac{16}{23}=\frac{140}{27}-\frac{5}{27}+\frac{7}{23}+\frac{16}{23}+\frac{1}{2}=\frac{135}{27}+\frac{23}{23}+\frac{1}{2}=5+1+\frac{1}{2}=\frac{13}{2}\)2,a,\(\frac{1}{2}+\frac{2}{3}.x=\frac{1}{4}\)
<=>\(\frac{2}{3}.x=-\frac{1}{2}\)
<=>\(x=-\frac{3}{4}\)
b,\(\frac{3}{5}+\frac{2}{5}\div x=3\frac{1}{2}\)
<=>\(\frac{2}{5x}=\frac{29}{10}\)
<=>\(x=\frac{29}{4}\)
c,\(\frac{x+4}{2004}+\frac{x+3}{2005}=\frac{x+2}{2006}+\frac{x+1}{2007}\)
<=> \(\frac{x+4}{2004}+1+\frac{x+3}{2005}+1=\frac{x+2}{2006}+1+\frac{x+1}{2007}+1\)
<=>\(\frac{x+2008}{2004}+\frac{x+2008}{2005}=\frac{x+2008}{2006}+\frac{x+2008}{2007}\)
<=>\(\left(x+2008\right)\left(\frac{1}{2004}+\frac{1}{2005}-\frac{1}{2006}-\frac{1}{2007}\right)\)=0
<=>x+2008=0 vì cái ngoặc còn lại\(\ne0\)
<=>x=-2008
Vậy x=-2008
Bạn nhớ tk cho mình vì mình đã chăm chỉ làm hết bài bạn hỏi nha!
a) ĐKXĐ : \(x\ne0\)
\(\left(-3+\frac{3}{x}-\frac{1}{3}\right):\left(1+\frac{2}{5}+\frac{2}{3}\right)=\frac{-5}{4}\)
\(\left(\frac{-9x}{3x}+\frac{9}{3x}-\frac{x}{3x}\right):\left(\frac{15}{15}+\frac{6}{15}+\frac{10}{15}\right)=\frac{-5}{4}\)
\(\frac{-9x+9-x}{3x}:\frac{15+6+10}{15}=\frac{-5}{4}\)
\(\frac{-10x+9}{3x}:\frac{31}{15}=\frac{-5}{4}\)
\(\frac{-10x+9}{3x}=\frac{-31}{12}\)
\(\Leftrightarrow12\left(-10x+9\right)=-31\cdot3x\)
\(\Leftrightarrow-120x+108=-93x\)
\(\Leftrightarrow-120x+93x=-108\)
\(\Leftrightarrow-27x=-108\)
\(\Leftrightarrow x=4\)
b) ĐKXĐ : \(x\ne0\)
\(\frac{-3x}{4}\cdot\left(\frac{1}{x}+\frac{2}{7}\right)=0\)
\(\Rightarrow\orbr{\begin{cases}\frac{-3x}{4}=0\\\frac{1}{x}+\frac{2}{7}=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=0\left(loai\right)\\\frac{-2}{-2x}=\frac{-2}{7}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=0\left(loai\right)\\x=\frac{-7}{2}\end{cases}}\)
Vậy.....
c) phân tích ra rồi làm thôi e :)) a bận rồi
Bài 1:
a) \(\left(\frac{9}{25}-2.18\right):\left(3\frac{4}{5}+0,2\right)\)
\(=\left(\frac{9}{25}-36\right):\left(\frac{19}{5}+\frac{1}{5}\right)\)
\(=\left(\frac{9}{25}-\frac{900}{25}\right):4\)
\(=-\frac{891}{25}.\frac{1}{4}\)
\(=-\frac{891}{100}\)
b) \(\frac{3}{8}.19\frac{1}{3}-\frac{3}{8}.33\frac{1}{3}\)
\(=\frac{3}{8}.\frac{58}{3}-\frac{3}{8}.\frac{100}{3}\)
\(=\frac{3}{8}\left(\frac{58}{3}-\frac{100}{3}\right)\)
\(=\frac{3}{8}\left(-\frac{42}{3}\right)\)
\(=\frac{3}{8}.\left(-14\right)\)
\(=-\frac{21}{4}\)
c) \(1\frac{4}{23}+\frac{5}{21}-\frac{4}{23}+0,5+\frac{16}{21}\)
\(=\frac{27}{23}+\frac{5}{21}-\frac{4}{23}+\frac{1}{2}+\frac{16}{21}\)
\(=\frac{27}{23}+\frac{5}{21}+\left(-\frac{4}{23}\right)+\frac{1}{2}+\frac{16}{21}\)
\(=\left[\frac{27}{23}+\left(-\frac{4}{23}\right)\right]+\left(\frac{5}{21}+\frac{16}{21}\right)+\frac{1}{2}\)
\(=1+1=2\)
d) \(\frac{21}{47}+\frac{9}{45}+\frac{26}{47}+\frac{4}{5}\)
\(=\frac{21}{47}+\frac{9}{45}+\frac{26}{47}+\frac{36}{45}\)
\(=\left(\frac{21}{47}+\frac{26}{47}\right)+\left(\frac{9}{45}+\frac{36}{45}\right)\)
\(=1+1=2\)
X = 0
còn cách tính thì mình ko biết