tìm x trong tỉ lệ thức sau :
a) \(2\frac{1}{3}\div\frac{1}{3}=\frac{7}{9}\div x\) b) \(x\div\frac{1}{3}=\frac{12}{99}\div\frac{15}{90}\)
c) \(\frac{4}{9}\div x=3\frac{1}{3}\div2,25\) d ) \(\frac{3}{4}\div\frac{41}{99}=x\div\frac{75}{90}\)
Cho và
Khi đó
Câu 6: \(\left|\dfrac{5}{6}x-10\right|\le0\)
mà \(\left|\dfrac{5}{6}x-10\right|\ge0\forall x\)
\(\Rightarrow\left|\dfrac{5}{6}x-10\right|=0\)
\(\Rightarrow\dfrac{5}{6}x-10=0\Rightarrow\dfrac{5}{6}x=10\)
\(\Rightarrow x=10:\dfrac{5}{6}=10.\dfrac{6}{5}=12\)
Vậy \(x=12\)
Câu 7:
\(2^{3x+2}=4^{x+5}\)
\(\Rightarrow2^{3x+2}=\left(2^2\right)^{x+5}\)
\(\Rightarrow2^{3x+2}=2^{2\left(x+5\right)}\)
\(\Rightarrow3x+2=2\left(x+5\right)\)
\(\Rightarrow3x+2=2x+10\)
\(\Rightarrow3x-2x=10-2\)
\(\Rightarrow x=8\)
Vậy \(x=8\)
Tính nhanh
a) \(\frac{2}{3}\div\frac{4}{9}+\frac{1}{3}\div\frac{4}{9}=\)
b) \(\frac{9}{8}\div\frac{5}{7}-\frac{1}{8}\div\frac{5}{7}=\)
c) \(\frac{2}{5}\div\frac{7}{3}-\frac{2}{5}\div\frac{1}{3}=\)
d) \(\frac{3}{7}\div\frac{7}{3}+\frac{3}{7}\div\frac{9}{4}=\)
a)\(\frac{2}{3}:\frac{4}{9}+\frac{1}{3}:\frac{4}{9}\)
=\(\frac{4}{9}:\left(\frac{2}{3}+\frac{1}{3}\right)\)
=\(\frac{4}{9}:1=\frac{4}{9}\)
b
Tìm x biết :
a/ \(\frac{62}{7}\cdot x=\frac{29}{9}\div\frac{3}{56}\)
b/ \(\frac{1}{5}\div x=\frac{1}{5}+\frac{1}{7}\)
c/ \(\frac{1}{2\cdot a^2+1}\div x=2\)
\(a)\frac{62}{7}\cdot x=\frac{29}{9}\div\frac{3}{56}\)
\(\Rightarrow\frac{62}{7}\cdot x=\frac{29}{9}\cdot\frac{56}{3}\)
\(\Rightarrow\frac{62}{7}\cdot x=\frac{1624}{27}\)
\(\Rightarrow x=\frac{1624}{27}\div\frac{62}{7}\)
\(\Rightarrow x=\frac{1624}{27}\cdot\frac{7}{62}\)
\(\Rightarrow x=\frac{11368}{1674}=\frac{5684}{837}\)
Rút gọn thử đi
\(1.\) Tìm x : \(\frac{x+11}{12}+\frac{x+11}{13}+\frac{x+11}{14}=\frac{x+11}{15}+\frac{x+11}{16}\)
Thay x tìm được để tính A = \(1+x+x^2+x^3+...+x^{99}+x^{100}\)
\(2.\) Tìm x ϵ Z để A ϵ Z và tìm giá trị đó\
a) \(A=\frac{x+3}{x+2}\) b) \(A=\frac{1-2x}{x+3}\)
\(3.\) Tìm min của : /x-2014/ + /x-2013/ + /x+2012/+/x+2011/ + /x+2010/ + 2014 P/s : +) Mk k pjt viết dấu GTTĐ trên đây nên mk viết tạm dấu "/" nha +) Mk đã làm dc r nhưng mk hỏi ý kiến mn cho đầy đủ nha THANKS TOO MUCH !!!!!!!!!!!!!!!!!!!!!!!!!
\(\frac{x+11}{12}+\frac{x+11}{13}+\frac{x+11}{14}=\frac{x+11}{15}+\frac{x+11}{16}\)
\(\Rightarrow\frac{x+11}{12}+\frac{x+11}{13}+\frac{x+11}{14}-\frac{x+11}{15}-\frac{x+11}{16}=0\)
\(\Rightarrow\left(x+11\right)\left(\frac{1}{12}+\frac{1}{13}+\frac{1}{14}-\frac{1}{15}-\frac{1}{16}\right)=0\)
Mà \(\left(\frac{1}{12}+\frac{1}{13}+\frac{1}{14}-\frac{1}{15}-\frac{1}{16}\right)\ne0\)
\(\Rightarrow x+11=0\Rightarrow x=-11\)
a)\(A=\frac{x+3}{x+2}=\frac{x+2+1}{x+2}=1+\frac{1}{x+2}\)
\(\Rightarrow A\in Z\Leftrightarrow1+\frac{1}{x+2}\in Z\Leftrightarrow\frac{1}{x+2}\in Z\Leftrightarrow1⋮x+2\Leftrightarrow x+2\inƯ\left(1\right)\)
\(\Rightarrow x+2\in\left\{-1;1\right\}\)
\(\Rightarrow x\in\left\{-3;-1\right\}\)
Nếu x=-3\(\Rightarrow A=\frac{-3+3}{-3+2}=\frac{0}{-1}=0\)
Nếu x=-1\(\Rightarrow A=\frac{-1+3}{-1+2}=\frac{2}{1}=2\)
tìm x biết
1, \(x\cdot\frac{6}{25}=\frac{15}{-13}\)
2, \(x\div\frac{4}{10}=\frac{13}{-45}+\frac{8}{15}\)
3, \(\frac{3}{8}-\frac{1}{6}\cdot x=\frac{1}{4}\)
4, \(\frac{1}{3}+\frac{1}{2}\div x=-4\)
5, \(x+\frac{7}{12}=\frac{17}{18}-\frac{1}{9}\)
\(x\times\frac{6}{25}=\frac{15}{-13}\)
x=\(\frac{15}{-13}\div\frac{6}{25}\)
x=\(-\frac{125}{26}\)
các câu còn lại làm tương tự nha!!!
\(1.x.\frac{6}{25}=\frac{15}{-13}\\ x=\frac{15}{-13}:\frac{6}{25}\\ x=-\frac{125}{26}\)
\(2.x:\frac{4}{10}=\frac{13}{-45}+\frac{8}{15}\\ x:\frac{4}{10}=\frac{11}{45}\\ x=\frac{11}{45}.\frac{4}{10}\\ x=\frac{22}{225}\)
\(3.\frac{3}{8}-\frac{1}{6}.x=\frac{1}{4}\\ \frac{1}{6}.x=\frac{3}{8}-\frac{1}{4}\\ \frac{1}{6}.x=\frac{1}{8}\\ x=\frac{1}{8}:\frac{1}{6}\\ x=\frac{3}{4}\)
\(4.\frac{1}{3}+\frac{1}{2}:x=-4\\ \frac{1}{2}:x=-4-\frac{1}{3}=-\frac{13}{3}\\ x=\frac{1}{2}:\left(-\frac{13}{3}\right)=-\frac{3}{26}\)
\(5.x+\frac{7}{12}=\frac{17}{18}-\frac{1}{9}=\frac{5}{6}\\ x=\frac{5}{6}-\frac{7}{12}\\ x=\frac{1}{4}\)
tìm x biết
1, \(x\cdot\frac{6}{25}=\frac{15}{-13}\)
2, \(x\div\frac{4}{10}=\frac{13}{-45}+\frac{8}{15}\)
3, \(\frac{3}{8}-\frac{1}{6}\cdot x=\frac{1}{4}\)
4, \(\frac{1}{3}+\frac{1}{2}\div x=-4\)
5, \(x+\frac{7}{12}=\frac{17}{18}-\frac{1}{9}\)
a) Áp dụng t/c dãy tỉ số bằng nhau, ta có:
\(\frac{x}{2}=\frac{y}{3}=\frac{y-x}{3-2}=\frac{14}{1}=14\)
=> \(\begin{cases}x=28\\y=42\end{cases}\)
b) Từ 2x = 7y => \(\frac{2x}{14}=\frac{7y}{14}\Rightarrow\frac{x}{7}=\frac{y}{2}\)
Áp dụng t/c dãy tỉ số bằng nhau, ta có:
\(\frac{x}{7}=\frac{y}{2}=\frac{x+y}{7+2}=\frac{36}{9}=4\)
=> \(\begin{cases}x=28\\y=8\end{cases}\)
c) Từ \(\frac{x}{y}=\frac{3}{7}\Rightarrow\frac{x}{7}=\frac{y}{3}\)
Áp dụng t/c dãy tỉ số bằng nhau, ta có:
\(\frac{x}{7}=\frac{y}{3}=\frac{y-x}{3-7}=\frac{20}{-4}=-5\)
=> \(\begin{cases}x=-35\\y=-15\end{cases}\)
d) Đặt \(\frac{x}{2}=\frac{y}{3}=k\Rightarrow\begin{cases}x=2k\\y=3k\end{cases}\)
Vì xy = 24 => 2k.3k = 24 => 6k2 = 24 => k2 = 4 => k = \(\pm\) 2
Với k = 2 => \(\begin{cases}x=4\\y=6\end{cases}\)
Với k = -2 => \(\begin{cases}x=-4\\y=-6\end{cases}\)
Bài 1 :Tìm x để :
\(\frac{x-2}{x-6}>0\)
Bài 2:Tim x de : \(\frac{3x+9}{x-4}\)nhận giá trị nguyên
Bài 3 : Thực hiện phép tính : \(\frac{1}{2}\div\left(-1\frac{3}{2}\right)\div1\frac{1}{3}\div\left(-1\frac{1}{4}\right)\div1\frac{1}{5}\div...\div1\frac{1}{99}\div\left(-1\frac{1}{100}\right)\)
Giúp mình với nhé. Mình tick cho.thank you
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#
\(\frac{1}{2}\div\left(-1\frac{1}{2}\right)\div1\frac{1}{3}\div\left(-1\frac{1}{4}\right)\div1\frac{1}{5}\div...\div1\frac{1}{99}\div\left(-1\frac{1}{100}\right)\)
\(=\frac{1}{2}\div\left(-\frac{3}{2}\right)\div\frac{4}{3}\div\left(-\frac{5}{4}\right)\div\frac{6}{5}\div...\div\frac{100}{99}\div\left(-\frac{101}{100}\right)\)
\(=\frac{1}{2}.\frac{-2}{3}.\frac{3}{4}.\frac{-4}{5}.\frac{5}{6}...\frac{99}{100}.\frac{-100}{101}\)
\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.\frac{4}{5}.\frac{5}{6}...\frac{99}{100}.\frac{100}{101}\) ( do có 50 thừa số âm )
\(=\frac{1.2.3.4.5...99.100}{2.3.4.5.6...100.101}\)
\(=\frac{1}{101}\)
Tìm x biết:
a)\(\frac{3}{5}\times x-\frac{1}{5}=\frac{4}{5}\)
b)\(\frac{4}{7}+\frac{5}{7}\div x=1\)
c)\(\frac{-12}{7}\times\left(\frac{3}{4}-x\right)\times\frac{1}{4}=-1\)
d)\(x\div\frac{17}{8}=\frac{-2}{5}\times\frac{-9}{17}+3\)
a) \(\frac{3}{5}.x-\frac{1}{5}=\frac{4}{5}\)
\(\Leftrightarrow\frac{3}{5}.x=\frac{4}{5}+\frac{1}{5}\)
\(\Leftrightarrow\frac{3}{5}.x=1\)
\(\Leftrightarrow x=1:\frac{3}{5}\)
\(\Leftrightarrow x=\frac{5}{3}\)
Vậy : \(x=\frac{5}{3}\)
b) \(\frac{4}{7}+\frac{5}{7}:x=1\)
\(\Leftrightarrow\frac{5}{7}:x=1-\frac{4}{7}\)
\(\Leftrightarrow\frac{5}{7}:x=\frac{3}{7}\)
\(\Leftrightarrow x=\frac{5}{7}:\frac{3}{7}\)
\(\Leftrightarrow x=\frac{5}{3}\)
Vậy : \(x=\frac{5}{3}\)
c) \(-\frac{12}{7}.\left(\frac{3}{4}-x\right).\frac{1}{4}=-1\)
\(\Leftrightarrow\frac{-12.1}{7.4}.\left(\frac{3}{4}-x\right)=-1\)
\(\Leftrightarrow-\frac{3}{7}.\left(\frac{3}{4}-x\right)=-1\)
\(\Leftrightarrow\frac{3}{4}-x=-1:\left(-\frac{3}{7}\right)\)
\(\Leftrightarrow\frac{3}{4}-x=\frac{7}{3}\)
\(\Leftrightarrow x=\frac{3}{4}-\frac{7}{3}=-\frac{19}{12}\)
Vậy : \(x=-\frac{19}{12}\)
d) \(x:\frac{17}{8}=-\frac{2}{5}.-\frac{9}{17}+3\)
\(\Leftrightarrow x:\frac{17}{8}=\frac{273}{85}\)
\(\Leftrightarrow x=\frac{273}{85}.\frac{17}{8}\)
\(\Leftrightarrow x=\frac{273}{40}\)
Vậy : \(x=\frac{273}{40}\)
\(\)