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giống cái kia thôi bn
Mik làm rồi mà
Mà cái bn Nguyễn Duy Đạt gì đó làm thiếu 1 trường hợp
Mà bn vẫn kik hở
Sao zzzzz??????
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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
\(c,\Rightarrow\left[{}\begin{matrix}-2\left(x+2\right)+\left(4-x\right)=11\left(x< -2\right)\\2\left(x+2\right)+\left(4-x\right)=11\left(-2\le x\le4\right)\\2\left(x+2\right)+\left(x-4\right)=11\left(x>4\right)\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{11}{3}\left(tm\right)\\x=3\left(tm\right)\\x=\dfrac{11}{3}\left(ktm\right)\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{11}{3}\end{matrix}\right.\)
\(a,\Rightarrow\left[{}\begin{matrix}x+\dfrac{5}{2}=3x+1\\x+\dfrac{5}{2}=-3x-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=-\dfrac{7}{8}\end{matrix}\right.\)
a/dễ --> tự lm
b/ \(\left(x-\dfrac{4}{7}\right)\left(1\dfrac{3}{5}+2x\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-\dfrac{1}{5}=0\\1\dfrac{3}{5}+2x=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}\\2x=\dfrac{8}{5}\Rightarrow x=\dfrac{4}{5}\end{matrix}\right.\)
Vậy...............
c/ \(\left(x-\dfrac{4}{7}\right):\left(x+\dfrac{1}{2}\right)>0\)
TH1: \(\left\{{}\begin{matrix}x-\dfrac{4}{7}>0\\x+\dfrac{1}{2}>0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x>\dfrac{4}{7}\\x>-\dfrac{1}{2}\end{matrix}\right.\) \(\Rightarrow x>\dfrac{4}{7}\)
TH2: \(\left\{{}\begin{matrix}x-\dfrac{4}{7}< 0\\x+\dfrac{1}{2}< 0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x< \dfrac{4}{7}\\x< -\dfrac{1}{2}\end{matrix}\right.\) \(\Rightarrow x< -\dfrac{1}{2}\)
Vậy \(x>\dfrac{4}{7}\) hoặc \(x< -\dfrac{1}{2}\) thì thỏa mãn đề
d/ \(\left(2x-3\right):\left(x+1\dfrac{3}{4}\right)< 0\)
TH1: \(\left\{{}\begin{matrix}2x-3>0\\x+1\dfrac{3}{4}< 0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x>1,5\\x< -\dfrac{7}{4}\end{matrix}\right.\)(vô lý)
TH2: \(\left\{{}\begin{matrix}2x-3< 0\\x+1\dfrac{3}{4}>0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x< 1,5\\x>-\dfrac{7}{4}\end{matrix}\right.\)\(\Rightarrow-\dfrac{7}{4}< x< 1,5\)
Vậy...................
a/ 2x - 10 - [3x - 14 - (4 - 5x) - 2x] = 2
=> 2x - 10 - (3x - 14 - 4 + 5x - 2x) = 2
=> 2x - 10 - 3x + 14 + 4 - 5x + 2x = 2
=> -4x + 6 = 0
=> -4x = -6
=> x = 3/2
b/ \(\left(\frac{1}{4}x-1\right)+\left(\frac{5}{6}x-2\right)-\left(\frac{3}{8}x+1\right)=4,5\)
\(\Rightarrow\frac{1}{4}x-1+\frac{5}{6}x-2-\frac{3}{8}x-1-\frac{9}{2}=0\)
\(\Rightarrow\frac{17}{24}x-\frac{17}{2}=0\)
\(\Rightarrow\frac{17}{24}x=\frac{17}{2}\)
\(\Rightarrow x=12\)
\(a,3-x=x+1,8\)
\(\Rightarrow-x-x=1,8-3\)
\(\Rightarrow-2x=-1,2\)
\(\Rightarrow x=0,6\)
\(b,2x-5=7x+35\)
\(\Rightarrow2x-7x=35+5\)
\(\Rightarrow-5x=40\)
\(\Rightarrow x=-8\)
\(c,2\left(x+10\right)=3\left(x-6\right)\)
\(\Rightarrow2x+20=3x-18\)
\(\Rightarrow2x-3x=-18-20\)
\(\Rightarrow-x=-38\)
\(\Rightarrow x=38\)
\(d,8\left(x-\dfrac{3}{8}\right)+1=6\left(\dfrac{1}{6}+x\right)+x\)
\(\Rightarrow8x-3+1=1+6x+x\)
\(\Rightarrow8x-3=7x\)
\(\Rightarrow8x-7x=3\)
\(\Rightarrow x=3\)
\(e,\dfrac{2}{9}-3x=\dfrac{4}{3}-x\)
\(\Rightarrow-3x+x=\dfrac{4}{3}-\dfrac{2}{9}\)
\(\Rightarrow-2x=\dfrac{10}{9}\)
\(\Rightarrow x=-\dfrac{5}{9}\)
\(g,\dfrac{1}{2}x+\dfrac{5}{6}=\dfrac{3}{4}x-\dfrac{1}{2}\)
\(\Rightarrow\dfrac{1}{2}x-\dfrac{3}{4}x=-\dfrac{1}{2}-\dfrac{5}{6}\)
\(\Rightarrow-\dfrac{1}{4}x=-\dfrac{4}{3}\)
\(\Rightarrow x=\dfrac{16}{3}\)
\(h,x-4=\dfrac{5}{6}\left(6-\dfrac{6}{5}x\right)\)
\(\Rightarrow x-4=5-x\)
\(\Rightarrow x+x=5+4\)
\(\Rightarrow2x=9\)
\(\Rightarrow x=\dfrac{9}{2}\)
\(k,7x^2-11=6x^2-2\)
\(\Rightarrow7x^2-6x^2=-2+11\)
\(\Rightarrow x^2=9\Rightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)
\(m,5\left(x+3\cdot2^3\right)=10^2\)
\(\Rightarrow5\left(x+24\right)=100\)
\(\Rightarrow x+24=20\)
\(\Rightarrow x=-4\)
\(n,\dfrac{4}{9}-\left(\dfrac{1}{6^2}\right)=\dfrac{2}{3}\left(x-\dfrac{2}{3}\right)^2+\dfrac{5}{12}\)
\(\Rightarrow\dfrac{2}{3}\left(x-\dfrac{2}{3}\right)^2+\dfrac{5}{12}=\dfrac{4}{9}-\dfrac{1}{36}\)
\(\Rightarrow\dfrac{2}{3}\left(x-\dfrac{2}{3}\right)^2+\dfrac{5}{12}=\dfrac{5}{12}\)
\(\Rightarrow\dfrac{2}{3}\left(x-\dfrac{2}{3}\right)^2=0\)
\(\Rightarrow x-\dfrac{2}{3}=0\Rightarrow x=\dfrac{2}{3}\)
#\(Urushi\text{☕}\)
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
Ta có:
a) \(\left(2x-1\right)^4=8\Leftrightarrow\left(2x-1\right)^4=2^4\)\(\Leftrightarrow\left[{}\begin{matrix}2x-1=2\\2x-1=-2\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{1}{2}\end{matrix}\right.\)
b) \(\left(x-1\right)^5=-32\Leftrightarrow\left(x-1\right)^5=-2^5\Leftrightarrow x-1=-2\Leftrightarrow x=-1\)
c) \(\left(2x-1\right)^6=\left(2x-1\right)^8\Leftrightarrow\left(2x-1\right)^6-\left(2x-1\right)^8\Leftrightarrow\left(2x-1\right)^6\left[1-\left(2x-1\right)^2\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(2x-1\right)^6=0\\1-\left(2x-1\right)^2=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\\left(2x-1\right)^2=1\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\2x-1=1\\2x-1=-1\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=1\\x=0\end{matrix}\right.\)
Câu a mình sửa lại nhé. Xin lỗi đã sai sót mong thông cảm ạ.
a) \(\left(2x-1\right)^4=8\Leftrightarrow\left(2x-1\right)^4=\left(\sqrt[4]{8}\right)^4\)\(\Leftrightarrow\left[{}\begin{matrix}2x-1=\sqrt[4]{8}\\2x-1=-\sqrt[4]{8}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=\sqrt[4]{8}+1\\2x=-\sqrt[4]{8}+1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\sqrt[4]{8}+1}{2}\\x=\dfrac{-\sqrt[4]{8}+1}{2}\end{matrix}\right.\)