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a) \(\left(x-1\right)\left(2x-4\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-1=0\Rightarrow x=1\\2x-4=0\Rightarrow x=2\end{matrix}\right.\)
b) \(\left(x^2+5\right)\left(x-5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x^2+5=0\Rightarrow x=-\sqrt{5}\\x-5=0\Rightarrow x=5\end{matrix}\right.\)
mà \(x\in Z\Rightarrow x=5\)
c) \(\left(x^2+5\right)\left(x^2-2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x^2+5=0\Rightarrow x=-\sqrt{5}\\x^2-2=0\Rightarrow x=\sqrt{2}\end{matrix}\right.\)
mà \(x\in Z\Rightarrow x\in\varnothing\)
1a) \(\left|\frac{3}{2}x+\frac{1}{2}\right|=\left|4x-1\right|\)
=> \(\orbr{\begin{cases}\frac{3}{2}x+\frac{1}{2}=4x-1\\\frac{3}{2}x+\frac{1}{2}=1-4x\end{cases}}\)
=> \(\orbr{\begin{cases}-\frac{5}{2}x=-\frac{3}{2}\\\frac{11}{2}x=\frac{1}{2}\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{5}{3}\\x=\frac{1}{11}\end{cases}}\)
b) \(\left|\frac{5}{4}x-\frac{7}{2}\right|-\left|\frac{5}{8}x+\frac{3}{5}\right|=0\)
=>\(\left|\frac{5}{4}x-\frac{7}{2}\right|=\left|\frac{5}{8}x+\frac{3}{5}\right|\)
=> \(\orbr{\begin{cases}\frac{5}{4}x-\frac{7}{2}=\frac{5}{8}x+\frac{3}{5}\\\frac{5}{4}x-\frac{7}{2}=-\frac{5}{8}x-\frac{3}{5}\end{cases}}\)
=> \(\orbr{\begin{cases}\frac{5}{8}x=\frac{41}{10}\\\frac{15}{8}x=\frac{29}{10}\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{164}{25}\\x=\frac{116}{75}\end{cases}}\)
c) TT
a, \(\left|\frac{3}{2}x+\frac{1}{2}\right|=\left|4x-1\right|\)
=> \(\orbr{\begin{cases}\frac{3}{2}x+\frac{1}{2}=4x-1\\-\frac{3}{2}x-\frac{1}{2}=4x-1\end{cases}}\)
=> \(\orbr{\begin{cases}\frac{3}{2}x+\frac{1}{2}-4x=-1\\-\frac{3}{2}x-\frac{1}{2}-4x=-1\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{3}{5}\\x=\frac{1}{11}\end{cases}}\)
\(b,\left|\frac{5}{4}x-\frac{7}{2}\right|-\left|\frac{5}{8}x+\frac{3}{5}\right|=0\)
=> \(\left|\frac{5}{4}x-\frac{7}{2}\right|-0=\left|\frac{5}{8}x+\frac{3}{5}\right|\)
=> \(\frac{\left|5x-14\right|}{4}=\frac{\left|25x+24\right|}{40}\)
=> \(\frac{10(\left|5x-14\right|)}{40}=\frac{\left|25x+24\right|}{40}\)
=> \(\left|50x-140\right|=\left|25x+24\right|\)
=> \(\orbr{\begin{cases}50x-140=25x+24\\-50x+140=25x+24\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{164}{25}\\x=\frac{116}{75}\end{cases}}\)
c, \(\left|\frac{7}{5}x+\frac{2}{3}\right|=\left|\frac{4}{3}x-\frac{1}{4}\right|\)
=> \(\orbr{\begin{cases}\frac{7}{5}x+\frac{2}{3}=\frac{4}{3}x-\frac{1}{4}\\-\frac{7}{5}x-\frac{2}{3}=\frac{4}{3}x-\frac{1}{4}\end{cases}}\)
=> \(\orbr{\begin{cases}x=-\frac{55}{4}\\x=-\frac{25}{164}\end{cases}}\)
Bài 2 : a. |2x - 5| = x + 1
TH1 : 2x - 5 = x + 1
=> 2x - 5 - x = 1
=> 2x - x - 5 = 1
=> 2x - x = 6
=> x = 6
TH2 : -2x + 5 = x + 1
=> -2x + 5 - x = 1
=> -2x - x + 5 = 1
=> -3x = -4
=> x = 4/3
Ba bài còn lại tương tự
1.
a) \(x\in\left\{4;5;6;7;8;9;10;11;12;13\right\}\)
b) x=0
d) \(x=\frac{-1}{35}\) hoặc \(x=\frac{-13}{35}\)
e) \(x=\frac{2}{3}\)
a, Ta có : \(\left(2x-1\right)^4=16\)
=> \(\left(\left(2x-1\right)^2\right)^2-\left(2^2\right)^2=0\)
=> \(\left(\left(2x-1\right)^2-2^2\right)\left(\left(2x-1\right)^2+2^2\right)=0\)
=> \(\left(2x-1-2\right)\left(2x-1+2\right)\left(\left(2x-1\right)^2+2^2\right)=0\)
Mà \(\left(2x-1\right)^2+2^2>0\)
=> \(\left(2x-3\right)\left(2x+1\right)=0\)
=> \(\left[{}\begin{matrix}x=\frac{3}{2}\\x=-\frac{1}{2}\end{matrix}\right.\)
Vậy phương trình có tập nghiệm là \(S=\left\{\frac{3}{2};-\frac{1}{2}\right\}\)
b, Ta có : \(\left(2x+1\right)^4=\left(2x+1\right)^6\)
=> \(\left(2x+1\right)^6-\left(2x+1\right)^4=0\)
=> \(\left(2x+1\right)^4\left(\left(2x+1\right)^2-1\right)=0\)
=> \(\left(2x+1\right)^4\left(2x+1-1\right)\left(2x+1+1\right)=0\)
=> \(2x\left(2x+1\right)^4\left(2x+2\right)=0\)
=> \(\left[{}\begin{matrix}2x=0\\2x+1=0\\2x+2=0\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=0\\x=-\frac{1}{2}\\x=-1\end{matrix}\right.\)
Vậy phương trình có tập nghiệm là \(S=\left\{0;-1;-\frac{1}{2}\right\}\)
c, Ta có : \(\left|\left|x+3\right|-8\right|=20\)
TH1 : \(x+3\ge0\left(x\ge-3\right)\)
=> \(\left|x+3\right|=x+3\)
=> \(\left|x-5\right|=20\)
TH1.1 : \(x-5\ge0\left(x\ge5\right)\)
=> \(\left|x-5\right|=x-5=20\)
=> \(x=25\left(TM\right)\)
TH1.2 : \(x-5< 0\left(x< 5\right)\)
=> \(\left|x-5\right|=5-x=20\)
=> \(x=-15\) ( không thỏa mãn )
TH2 : \(x+3< 0\left(x< -3\right)\)
=> \(\left|x+3\right|=-x-3\)
=> \(\left|-x-11\right|=20\)
TH1.1 : \(-x-11\ge0\left(x\le-11\right)\)
=> \(\left|-x-11\right|=-x-11=20\)
=> \(x=-31\left(TM\right)\)
TH1.2 : \(-x-11< 0\left(x>-11\right)\)
=> \(\left|-x-11\right|=x+11=20\)
=> \(x=9\) ( không thỏa mãn )
Vậy phương trình có tập nghiệm là \(S=\left\{-31;25\right\}\)
a, ( 2x - 1 )4 = 16
=> 2x - 1 = 2 hoặc -2
TH1: 2x - 1 = 2
=> 2x = 2 + 1 = 3; => x = \(\frac{3}{2}\)
TH2: 2x - 1 = -2
=> 2x = -2 + 1 = -1; => x =- \(\frac{1}{2}\)
b, ( 2x + 1 )4 = ( 2x + 1 )6
=> ( 2x + 1 )4 - ( 2x + 1 )6 = 0
= ( 2x + 1 )4 - ( 2x - 1 )2 . ( 2x - 1 )4
= ( 2x + 1 )4 . [ 1 - ( 2x - 1 )2 ] = 0
Ta có ( 2x + 1 )4 và ( 2x - 1 )2 \(\ge\) 0 vì có số mũ chẵn
Ta có 2 TH
TH1: ( 2x - 1 )4 = 0
=> 2x - 1 = 0; => x = \(\frac{1}{2}\)
TH2: 1 - ( 2x - 1 )2 = 0; => ( 2x - 1 )2 = 1
=> 2x - 1 = 1; => x = 1
c, //x + 3/ - 8/ = 20
Ta có 2 TH, mỗi TH lại chia thành 2 TH nhỏ hơn
TH1: /x + 3/ - 8 = 20
=> /x + 3/ = 28
=> x + 3 = 28 hoặc -28
TH1 nhỏ: x + 3 = 28; => x = 25
TH2 nhỏ: x + 3 = -28; => x = -31
TH2: /x + 3/ - 8 = -20
=> /x + 3/ = -12; => TH này loại
=> x = 25; -31
a)Áp dụng bđt \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\) ta có:
\(\left|x-1\right|+\left|3+x\right|=\left|1-x\right|+\left|3+x\right|\ge\left|1-x+3+x\right|=4\)
\(\Rightarrow VT\ge VP."="\Leftrightarrow-3\le x\le1\)
b) \(\hept{\begin{cases}\left|2x+3\right|+\left|2x-1\right|=\left|2x+3\right|+\left|1-2x\right|\ge4\\\frac{8}{2\left(y-5\right)^2+2}\le4\end{cases}}\Leftrightarrow VT\ge VP."="\Leftrightarrow\hept{\begin{cases}-\frac{3}{2}\le x\le\frac{1}{2}\\y=5\end{cases}}\)
c Tương tự b
2) \(\frac{1}{x}+\frac{1}{y}=5\Leftrightarrow x+y-5xy=0\Leftrightarrow5x+5y-25xy=0\Leftrightarrow5x\left(1-5y\right)-\left(1-5y\right)=-1\)
\(\Leftrightarrow\left(5x-1\right)\left(1-5y\right)=-1\)
Xét ước
a) => (2x -1)6 - (2x - 1)8 = 0 => (2x - 1)6 - (2x - 1)6.(2x - 1)2 = 0
=> (2x - 1)6.[1 - (2x - 1)2] = 0 => (2x - 1)6 = 0 hoặc 1 - (2x - 1)2 = 0
+) (2x - 1)6 = 0 => 2x - 1 = 0 => x = 1/2
+) 1 - (2x - 1)2 = 0 => (2x - 1)2 = 1 => 2x - 1 = 1 hoặc 2x - 1 = - 1
2x - 1 = 1 => x = 1
2x - 1 = - 1 => x = 0
Vậy x = 0 ; x = 1/2; x = 1
b) => 5x + 5.5x = 650
=> 6.5x = 650 => 5x = 650 : 6 = 325/3 => không có số x nào thỏa mãn
Vậy............