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b) \(\left(\frac{2}{3}x-1\right).\left(\frac{3}{4}x+\frac{1}{2}\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}\frac{2}{3}x-1=0\\\frac{3}{4}x+\frac{1}{2}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\frac{2}{3}x=1\\\frac{3}{4}x=-\frac{1}{2}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1:\frac{2}{3}\\x=\left(-\frac{1}{2}\right):\frac{3}{4}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{3}{2}\\x=-\frac{2}{3}\end{matrix}\right.\)
Vậy \(x\in\left\{\frac{3}{2};-\frac{2}{3}\right\}.\)
c) \(x:\frac{9}{14}=\frac{7}{3}:x\)
\(\Rightarrow\frac{x}{\frac{19}{4}}=\frac{\frac{7}{3}}{x}\)
\(\Rightarrow x.x=\frac{7}{3}.\frac{19}{4}\)
\(\Rightarrow x.x=\frac{133}{12}\)
\(\Rightarrow x^2=\frac{133}{12}\)
\(\Rightarrow\left[{}\begin{matrix}x=\sqrt{\frac{133}{12}}\\x=-\sqrt{\frac{133}{12}}\end{matrix}\right.\)
Vậy \(x\in\left\{\sqrt{\frac{133}{12}};-\sqrt{\frac{133}{12}}\right\}.\)
d) \(\left(3x-1\right)^{10}=\left(3x-1\right)^{20}\)
\(\Rightarrow\left(3x-1\right)^{10}-\left(3x-1\right)^{20}=0\)
\(\Rightarrow\left(3x-1\right)^{10}.\left[1-\left(3x-1\right)^{10}\right]=0\)
\(\Rightarrow\left[{}\begin{matrix}\left(3x-1\right)^{10}=0\\1-\left(3x-1\right)^{10}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}3x-1=0\\\left(3x-1\right)^{10}=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}3x=1\\3x-1=\pm1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=1:3\\3x-1=1\\3x-1=-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{1}{3}\\3x=2\\3x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{1}{3}\\x=\frac{2}{3}\\x=0\end{matrix}\right.\)
Vậy \(x\in\left\{\frac{1}{3};\frac{2}{3};0\right\}.\)
Chúc bạn học tốt!
a) \(\frac{7-8x}{6}=\frac{-4+2x}{5}\)
=> \(\left(7-8x\right).5=6\left(-4+2x\right)\)
=> 35 - 40x = -24 + 12x
=> 35 + 24 = 12x + 40x
=> 52x = 59
=> x = 59/52
b) \(\frac{1-3:x}{8}=\frac{8}{1-3:x}\)
=> (1 - 3: x)2 = 82
=> \(\orbr{\begin{cases}1-3:x=8\\1-3:x=-8\end{cases}}\)
=> \(\orbr{\begin{cases}3:x=-7\\3:x=9\end{cases}}\)
=> \(\orbr{\begin{cases}x=-\frac{3}{7}\\x=\frac{1}{3}\end{cases}}\)
c) \(\left(x+1\right)\left(x-2\right)\ge0\)
=> \(\hept{\begin{cases}x+1\ge0\\x-2\ge0\end{cases}}\)hoặc \(\hept{\begin{cases}x+1\le0\\x-2\le0\end{cases}}\)
=> \(\hept{\begin{cases}x\ge-1\\x\ge2\end{cases}}\) hoặc \(\hept{\begin{cases}x\le-1\\x\le2\end{cases}}\)
=> \(-1\le x\le2\)
h) \(\left(x+1\right)\left(x-3\right)\le0\)
=> \(\hept{\begin{cases}x+1\ge0\\x-3\le0\end{cases}}\)hoặc \(\hept{\begin{cases}x+1\le0\\x-3\ge0\end{cases}}\)
=> \(\hept{\begin{cases}x\ge-1\\x\le3\end{cases}}\) hoặc \(\hept{\begin{cases}x\le-1\\x\ge3\end{cases}}\) (loại)
= \(-1\le x\le3\)
1. A = 75(42004 + 42003 +...+ 42 + 4 + 1) + 25
A = 25 . [3 . (42004 + 42003 +...+ 42 + 4 + 1) + 1]
A = 25 . (3 . 42004 + 3 . 42003 +...+ 3 . 42 + 3 . 4 + 3 + 1)
A = 25 . (3 . 42004 + 3 . 42003 +...+ 3 . 42 + 3 . 4 + 4)
A = 25 . 4 . (3 . 42003 + 3 . 42002 +...+ 3 . 4 + 3 + 1)
A =100 . (3 . 42003 + 3 . 42002 +...+ 3 . 4 + 3 + 1) \(⋮\) 100
Mình làm cho bạn 2 câu khó hơn còn mấy câu còn lại dungf phương pháp quy đồng rồi chuyển vế là tính được mà
c, <=> [(x-1)/2009 ]-1 +[ (x-2)/2008] -1 = [(x-3)/2007]-1 +[(x-4)/2006]-1
<=> (x-2010)/2009 + (x-2010)/2008 = (x-2010)/2007 + (x-2010)/2006
<=> (x-2010)*(1/2009+1/2008-1/2007-1/2006)=0
=> x-2010=0 => x=2010
d, TH1 : cả hai cùng âm
=>> 2X-4 <O => X< 2
Và 9-3x<0 =>> x> 3
=>> loại
Th2 cả hai cùng dương
2x-4>O => x>2
Và 9-3x>O => x<3
=>> 2<x<3 (tm)
cau a dau nhi cuoi cung k phai j dau nha ! mk an lom !
\(a,\)\(\left|x+5\right|=\frac{1}{7}-\left|\frac{4}{3}-\frac{1}{6}\right|\)
\(\Leftrightarrow\left|x+5\right|=\frac{1}{7}-\frac{7}{6}\)
\(\Leftrightarrow\left|x+5\right|=\frac{-43}{42}\)
ta có |x+5| \(\ge\)0 \(\forall x\)
Mà \(-\frac{43}{42}< 0\)nên ko có giá trị x thoả mãn
b,
\(\left|x+\frac{2}{3}\right|=\frac{1}{2}-\left(\frac{1}{4}+\frac{2}{3}\right)\)
\(\Leftrightarrow\left|x+\frac{2}{3}\right|=\frac{11}{12}\)
\(\Leftrightarrow\orbr{\begin{cases}x+\frac{2}{3}=\frac{11}{12}\forall x\ge-\frac{2}{3}\\-x-\frac{2}{3}=\frac{11}{12}\forall< -\frac{2}{3}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{4}\\x=-\frac{19}{12}\end{cases}}\)(thoả mãn đk)
a) \(x:\left(-\frac{1}{3}\right)^3=-\frac{1}{3}\)
\(\Rightarrow x=\left(-\frac{1}{3}\right).\left(-\frac{1}{3}\right)^3\)
\(\Rightarrow x=\left(-\frac{1}{3}\right)^4\)
\(\Rightarrow x=\frac{1}{81}\)
Vậy \(x=\frac{1}{81}.\)
b) \(\frac{3}{4}:\frac{41}{99}=x:\frac{75}{90}\)
\(\Rightarrow\frac{297}{164}=x:\frac{75}{90}\)
\(\Rightarrow x=\frac{297}{164}.\frac{75}{90}\)
\(\Rightarrow x=\frac{495}{328}\)
Vậy \(x=\frac{495}{328}.\)
c) \(x+\left|-\frac{1}{2}\right|=3\frac{1}{3}-4\frac{1}{2}\)
\(\Rightarrow x+\frac{1}{2}=\frac{10}{3}-\frac{9}{2}\)
\(\Rightarrow x+\frac{1}{2}=-\frac{7}{6}\)
\(\Rightarrow x=\left(-\frac{7}{6}\right)-\frac{1}{2}\)
\(\Rightarrow x=-\frac{5}{3}\)
Vậy \(x=-\frac{5}{3}.\)
Chúc bạn học tốt!
ths bạn