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Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời. 1. a, 3x + |x - 2| = 8 2. 5.(x - 2) - 4.(1 - 3x) = |3 - 7| + 2.(1 + 2x) a ) ( x + 1 ) x ( x2 - 4 ) = 0 vậy chắc chắn 1 biểu thức phải bằng 0 để có kết quả đúng . vậy chỉ có thể là x2 - 4 = 0 vì phép còn lại là x + 1 = số nguyên dương x2 - 4 = 0 x = 2 b ) x15 = x vậy quá rõ x = 1 , 0 vì chỉ có 2 số này nhân bao nhiêu lần chính nó cũng bằng nó c ) ( x - 5 ) 4 = ( x - 5 )6 4 x - 625 = 6 x - 15625 4 x + 15625 - 625 = 6 x 4 x + 15000 = 6 x 15000 = 2 x x = 7500 d ) làm sau a. \(\left(x+1\right)\left(x^2-4\right)=0\) \(\Leftrightarrow\left(x+1\right)\left(x-2\right)\left(x+2\right)=0\) TH1: \(x+1=0\Rightarrow x=-1\) TH2: \(x-2=0\Rightarrow x=2\) TH3: \(x+2=0\Rightarrow x=-2\) Vậy:... b) \(x^{15}=x\) \(\Rightarrow x\in\left\{0;1;-1\right\}\) c) \(\left(x-5\right)^4=\left(x-5\right)^6\) TH1:\(x-5=1\Rightarrow x=6\) TH2: \(x-5=-1\Rightarrow x=4\) TH3: \(x-5=0\Rightarrow x=5\) d) \(\left(2x+1\right)^3=125\) \(\Leftrightarrow2x+1=\sqrt[3]{125}=5\) \(\Leftrightarrow x=2\) a, \(x^2-9=0\Rightarrow x^2=9\Rightarrow x\pm3\) b, \(\left(x-3\right)^2-25=0\Rightarrow\left(x-3\right)^2=25\) \(\Rightarrow\left\{{}\begin{matrix}x-3=5\\x-3=-5\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=8\\x=-2\end{matrix}\right.\) c, \(\left(x-3\right)\left(2x-5\right)=0\) \(\Rightarrow\left[{}\begin{matrix}x-3=0\\2x-5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\2x=5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{5}{2}\end{matrix}\right.\) d, \(\left(x-3\right)x-2\left(x-3\right)=0\) \(\Rightarrow\left(x-3\right)\left(x-2\right)=0\) \(\Rightarrow\left[{}\begin{matrix}x-3=0\\x-2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\x=2\end{matrix}\right.\) e, \(3x\left(x-1\right)-5\left(1-x\right)=0\) \(\Rightarrow3x\left(x-1\right)+5\left(x-1\right)=0\) \(\Rightarrow\left(x-1\right)\left(3x+5\right)=0\) \(\Rightarrow\left[{}\begin{matrix}x-1=0\\3x+5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{5}{3}\end{matrix}\right.\) g, \(x^2+6x-7=0\) \(\Rightarrow x^2-x+7x-7=0\) \(\Rightarrow x.\left(x-1\right)+7.\left(x-1\right)=0\) \(\Rightarrow\left(x-1\right)\left(x+7\right)=0\) \(\Rightarrow\left[{}\begin{matrix}x-1=0\\x+7=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=-7\end{matrix}\right.\) h,\(2x^2+5x-7=0\) \(\Rightarrow2x^2-2x+7x-7=0\) \(\Rightarrow2x.\left(x-1\right)+7.\left(x-1\right)=0\) \(\Rightarrow\left(x-1\right)\left(2x+7\right)=0\) \(\Rightarrow\left[{}\begin{matrix}x-1=0\\2x+7=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{7}{2}\end{matrix}\right.\) Chúc bạn học tốt!!! a) \(x^2-9=0\Leftrightarrow x^2=9\Leftrightarrow\left\{{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\) vậy \(x=3;x=-3\) b) \(\left(x-3\right)^2-25=0\Leftrightarrow\left(x-3\right)^2=25\Leftrightarrow\left\{{}\begin{matrix}x-3=5\\x-3=-5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=8\\x=-2\end{matrix}\right.\) vậy \(x=8;x=-2\) c) \(\left(x-3\right)\left(2x-5\right)=0\Leftrightarrow\left\{{}\begin{matrix}x-3=0\\2x-5=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=3\\x=\dfrac{5}{2}\end{matrix}\right.\) vậy \(x=3;x=\dfrac{5}{2}\) d)\(\left(x-3\right).x-2\left(x-3\right)=0\Leftrightarrow\left(x-2\right)\left(x-3\right)=0\) \(\Leftrightarrow\left\{{}\begin{matrix}x-2=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\x=3\end{matrix}\right.\) vậy \(x=2;x=3\) e) \(3x\left(x-1\right)-5\left(1-x\right)=0\Leftrightarrow\left(3x+5\right)\left(x-1\right)=0\) \(\Leftrightarrow\left\{{}\begin{matrix}3x+5=0\\x-1=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{-5}{3}\\x=1\end{matrix}\right.\) vậy \(x=\dfrac{-5}{3};x=1\) câu e t thấy sai sai nhưng vẫn làm ; bn coi lại đề nha g) \(x^2+6x-7=0\Leftrightarrow x^2-x+7x-7=0\) \(\Leftrightarrow x\left(x-1\right)+7\left(x-1\right)=0\Leftrightarrow\left(x+7\right)\left(x-1\right)=0\) \(\Leftrightarrow\left\{{}\begin{matrix}x+7=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-7\\x=1\end{matrix}\right.\) vậy \(x=-7;x=1\) h) \(2x^2+5x-7=0\Leftrightarrow2x^2-2x+7x-7=0\) \(\Leftrightarrow2x\left(x-1\right)+7\left(x-1\right)=0\Leftrightarrow\left(2x+7\right)\left(x-1\right)=0\) \(\Leftrightarrow\left\{{}\begin{matrix}2x+7=0\\x-1=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{-7}{2}\\x=1\end{matrix}\right.\) vậy \(x=\dfrac{-7}{2};x=1\) Bài 1: a, \(\left(x-2\right)^2=9\) \(\Rightarrow x-2\in\left\{-3;3\right\}\Rightarrow x\in\left\{-1;5\right\}\) b, \(\left(3x-1\right)^3=-8\) \(\Rightarrow3x-1=-2\Rightarrow3x=-1\) \(\Rightarrow x=-\dfrac{1}{3}\) c, \(\left(x+\dfrac{1}{2}\right)^2=\dfrac{1}{16}\) \(\Rightarrow x+\dfrac{1}{2}\in\left\{-\dfrac{1}{4};\dfrac{1}{4}\right\}\) \(\Rightarrow x\in\left\{-\dfrac{3}{4};-\dfrac{1}{4}\right\}\) d, \(\left(\dfrac{2}{3}\right)^x=\dfrac{4}{9}\) \(\Rightarrow\left(\dfrac{2}{3}\right)^x=\left(\dfrac{2}{3}\right)^2\) Vì \(\dfrac{2}{3}\ne\pm1;\dfrac{2}{3}\ne0\) nên \(x=2\) e, \(\left(\dfrac{1}{2}\right)^{x-1}=\dfrac{1}{16}\) \(\Rightarrow\left(\dfrac{1}{2}\right)^{x-1}=\left(\dfrac{1}{2}\right)^4\) Vì \(\dfrac{1}{2}\ne\pm1;\dfrac{1}{2}\ne0\) nên \(x-1=4\Rightarrow x=5\) 4a) \(\frac{-2}{3}x=\frac{3}{10}-\frac{1}{5}=\frac{1}{10}\) \(\Leftrightarrow x=\frac{1}{10}:\frac{-2}{3}=\frac{1}{10}.\frac{3}{-2}=\frac{3}{-20}\) Vậy x=\(\frac{3}{-20}\) b) \(\frac{2}{3}x-\frac{3}{2}x=\frac{5}{12}\) \(\Leftrightarrow\left(\frac{2}{3}-\frac{3}{2}\right)x=\frac{5}{12}\) \(\Leftrightarrow\frac{-5}{6}x=\frac{5}{12}\) \(\Leftrightarrow x=\frac{5}{12}:\frac{-5}{6}=\frac{5}{12}.\frac{6}{-5}=\frac{1}{-2}\) Vậy x=\(\frac{1}{-2}\) g)Sửa đề: \(\left|4x-1\right|=\left(-3\right)^2\) \(\Leftrightarrow\left|4x-1\right|=9\) \(\Rightarrow\left[{}\begin{matrix}4x-1=9\\4x-1=\left(-9\right)\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\frac{5}{2}\\x=-2\end{matrix}\right.\) Vậy \(x\in\left\{\frac{5}{2};-2\right\}\) i) \(\left(x-1^3\right)=125\) \(\Leftrightarrow x-1=125\) \(\Leftrightarrow x=125+1=126\) Vậy x=126 k) \(\left(x+\frac{1}{2}\right).\left(\frac{2}{3}-2x\right)=0\) \(\Rightarrow\left[{}\begin{matrix}x+\frac{1}{2}=0\\\frac{2}{3}-2x=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\frac{-1}{2}\\x=\frac{1}{3}\end{matrix}\right.\) Vậy \(x\in\left\{\frac{-1}{2};\frac{1}{3}\right\}\) 1, Ta có : a . 81 = 34 => 3x= 34 => x = 4 . b. 125 = 53 => 5x+2 = 53 =>x + 2 = 3 => x = 1 c. 23 * 2x - 1 = 64 => 23 + ( x - 1 ) = 64 = 26 => 3 + ( x - 1 ) = 6 => x - 1 = 6 - 3 = 3 x = 3 + 1 x = 4 1. A = (-2)(-3) - 5.|-5| + 125.\(\left(-\dfrac{1}{5}\right)^2\)
K
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NL
Nguyễn Lê Phước Thịnh
12 tháng 10 2023
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Các câu hỏi dưới đây có thể giống với câu hỏi trên
KL
3 tháng 8 2017
<=> |x - 2| = 8 - 3x
Xét 2 TH :
TH1: x - 2 = 8 - 3x
<=> x + 3x = 8 + 2
<=> 4x = 10
<=> x = \(\dfrac{5}{2}\) (thỏa mãn)
TH2: x - 2 = -(8 - 3x)
<=> x - 2 = -8 + 3x
<=> -2 + 8 = 3x - x
<=> 6 = 2x
<=> x = 3 (thỏa mãn)
b, 5 - |x - 1| = 4
<=> |x - 1| = 1
<=> \(\left[{}\begin{matrix}x-1=1\\x-1=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=0\end{matrix}\right.\) (thỏa mãn)
@Nguyễn Hoàng Vũ
KL
3 tháng 8 2017
<=> 5x - 10 - 4 + 12x = 4 + 2 + 4x
<=> 17x - 14 = 6 + 4x
<=> 17x - 4x = 6 + 14
<=> 13x = 20
<=> x = \(\dfrac{20}{13}\) (thỏa mãn)
@Nguyễn Hoàng Vũ
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KL
1 tháng 8 2017
= 6 - 25 + 125.\(\dfrac{1}{25}\)
= -19 + 5
= -14
@Shine Anna
