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Nguyễn Khánh Phương
Bài 1 :
a) 149 - ( 35 : x + 3 ) x 17 = 13
( 35 : x + 3 ) x 17 = 149 - 13
( 35 : x + 3 ) x 17 = 136
( 35 : x + 3 ) = 136 : 17
( 35 : x + 3 ) = 8
35 - x = 8 - 3
35 - x = 5
x = 35 - 5
x = 30
b, 121 : 11 − ( 4x + 5 ) : 3 = 4
11 − 4x + 5 : 3 = 4
4x + 5 : 3 = 11 − 4
4x + 5 : 3 = 7
4x + 5 = 7 x 3
4x + 5 = 21
4x = 21 − 5
4x = 16
x = 16 : 4
x = 4
5x . 5x + 1 . 7 = 875
\(\Rightarrow\) 52x + 1 = 875 : 7 = 125
\(\Rightarrow\) 52x + 1 = 53
\(\Rightarrow\) 2x + 1 = 3
\(\Rightarrow\) 2x = 2
\(\Rightarrow\) x = 1
5 + 3x + 1 = 86
\(\Rightarrow\) 3x + 1 = 81
\(\Rightarrow\) 3x + 1 = (\(\pm\)3)4
\(\Rightarrow\) x + 1 = \(\pm\)3
\(\Rightarrow\) x = 2, -4
1. \(6x^3-8=40\\ 6x^3=48\\ x^3=8\\ \Rightarrow x=2\)Vậy x = 2
2. \(4x^5+15=47\\ 4x^5=32\\ x^5=8\\ \Rightarrow x\in\varnothing\left(\text{vì }x\in N\right)\)Vậy x ∈ ∅
3. \(2x^3-4=12\\ 2x^3=16\\ x^3=8\\ \Rightarrow x=2\)Vậy x = 2
4. \(5x^3-5=0\\ 5x^3=5\\ x^3=1\\ \Rightarrow x=1\)Vậy x = 1
5. \(\left(x-5\right)^{2016}=\left(x-5\right)^{2018}\\ \Rightarrow\left[{}\begin{matrix}x-5=0\\x-5=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=5\\x=6\end{matrix}\right.\)Vậy \(x\in\left\{5;6\right\}\)
6. \(\left(3x-2\right)^{20}=\left(3x-1\right)^{20}\\ \Rightarrow3x-2=3x-1\\ 3x-3x=2-1\\ 0=1\left(\text{vô lí}\right)\)Vậy x ∈ ∅
7. \(\left(3x-1\right)^{10}=\left(3x-1\right)^{20}\\ \left(3x-1\right)^{10}=\left[\left(3x-1\right)^2\right]^{10}\\ \Rightarrow\left(3x-1\right)^2=3x-1\\ \left(3x-1\right)^2-\left(3x-1\right)=0\\ \left(3x-1\right)\left[\left(3x-1\right)-1\right]=0\\ \left(3x-1\right)\left(3x-2\right)=0\\ \Rightarrow\left[{}\begin{matrix}3x-1=0\\3x-2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}3x=1\\3x=2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{1}{3}\left(\text{loại vì }x\in N\right)\\x=\frac{2}{3}\left(\text{loại vì }x\in N\right)\end{matrix}\right.\)Vậy x ∈ ∅
8. \(\left(2x-1\right)^{50}=2x-1\\ \left(2x-1\right)^{50}-\left(2x-1\right)=0\\ \left(2x-1\right)\left[\left(2x-1\right)^{49}-1\right]=0\\ \Rightarrow\left[{}\begin{matrix}2x-1=0\\\left(2x-1\right)^{49}=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=1\\2x-1=1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\frac{1}{2}\left(\text{loại vì }x\in N\right)\\x=1\left(t/m\right)\end{matrix}\right.\)Vậy x = 1
9. \(\left(\frac{x}{3}-5\right)^{2000}=\left(\frac{x}{3}-5\right)^{2008}\\ \left(\frac{x}{3}-5\right)^{2008}-\left(\frac{x}{3}-5\right)^{2000}=0\\ \left(\frac{x}{3}-5\right)^{2000}\left[\left(\frac{x}{3}-5\right)^8-1\right]=0\\ \Rightarrow\left[{}\begin{matrix}\left(\frac{x}{3}-5\right)^{2000}=0\\\left(\frac{x}{3}-5\right)^8=1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}\frac{x}{3}-5=0\\\frac{x}{3}-5=1\\\frac{x}{3}-5=-1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}\frac{x}{3}=5\\\frac{x}{3}=6\\\frac{x}{3}=4\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=5\cdot3=15\\x=6\cdot3=18\\x=4\cdot3=12\end{matrix}\right.\)Vậy \(x\in\left\{15;18;12\right\}\)
\(1.6x^3-8=40\\ \Leftrightarrow6x^3=48\\ \Leftrightarrow x^3=8\Leftrightarrow x^3=2^3=\left(-2\right)^3\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)
Vậy \(x\in\left\{2;-2\right\}\)
\(2.4x^3+15=47\) (T nghĩ đề là mũ 3)
\(\Leftrightarrow4x^3=32\Leftrightarrow x^3=8=2^3=\left(-2\right)^3\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)
Vậy \(x\in\left\{2;-2\right\}\)
Câu 3, 4 tương tự nhé.
Bài 1:
1; 5\(x\) + \(x\) = 39 - 311 : 39
\(x\).(5 + 1) = 39 - 32
\(x.6\) = 39 - 9
\(x.6\) = 30
\(x\) = 30 : 6
\(x\) = 5
Vậy \(x\) = 5
2; 5\(x\) + \(x\) = 150 : 2 + 3
\(x\).(5 + 1) = 75 + 3
\(x.6\) = 78
\(x\) = 78 : 6
\(x\) = 13
Vậy \(x=13\)
\(a,\left(-5\right).\left|x\right|=-75\)
\(\left|x\right|=\frac{-75}{-5}=15\)
\(\Rightarrow\orbr{\begin{cases}x=15\\x=-15\end{cases}}\)
Vậy....
\(b,\left(-6\right)^3.x^2=-1944\)
\(-216.x^2=-1944\)
\(x^2=9\)
\(\Rightarrow x=\pm3\)
Vậy....
\(d,\left|9-x\right|=-7+64\)
\(\left|9-x\right|=57\)
\(\Rightarrow\orbr{\begin{cases}9-x=57\\9-x=-57\end{cases}\Rightarrow\orbr{\begin{cases}x=-48\\x=66\end{cases}}}\)
Vậy...
\(e,\left|x+101\right|-\left(-16\right)=\left(-43\right).\left(-5\right)\)
\(\left|x+101\right|+16=215\)
\(\left|x+101\right|=199\)
\(\Rightarrow\orbr{\begin{cases}x+101=199\\x+101=-199\end{cases}\Rightarrow\orbr{\begin{cases}x=98\\x=-300\end{cases}}}\)
Vậy..
hok tốt!!
a,\(\left(-5\right).\left|x\right|=-75\)
\(=>\left|x\right|=-75:\left(-5\right)=15\)
\(=>\orbr{\begin{cases}x=15\\x=-15\end{cases}}\)
b,\(\left(-6\right)^3.x^2=-1944\)
\(=>\frac{1944}{216}=x^2\)
\(=>x=\sqrt{\frac{1944}{216}}=3\)
a) \(\frac{3}{7}-\frac{1}{7}x=\frac{2}{3}\)
=> \(\frac{1}{7}x=\frac{3}{7}-\frac{2}{3}=-\frac{5}{21}\)
=> \(x=-\frac{5}{21}:\frac{1}{7}=-\frac{5}{21}\cdot7=-\frac{5}{3}\)
b) \(3x^2-2=72\)=> 3x2 = 74 => x2 = 74/3 => x không thỏa mãn
c) \(\left(19x+2\cdot5^2\right):14=\left(13-8\right)^2-4^2\)
=> \(\left(19x+2\cdot25\right):14=5^2-4^2=9\)
=> \(\left(19x+50\right):14=9\)
=> \(19x+50=126\)
=> \(19x=76\)
=> x = 4
d) \(x:\frac{1}{2}+x:\frac{1}{4}+x:\frac{1}{8}+x:\frac{1}{16}+x:\frac{1}{32}=343\)
=> \(x\cdot2+x\cdot4+x\cdot8+x\cdot16+x\cdot32=343\)
=> \(x\left(2+4+8+16+32\right)=343\)
=> x . 62 = 343
=> x = 343/62