Tìm số nguyên \(x\) biết:
a) \(\left|x-3\right|-\left(-3\right)=4\)
b) \(x-\left(1-x\right)=5+\left(-1+x\right)\)
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
4 tháng 10 2021
a: Ta có: \(\left(5x+1\right)^2-\left(5x-3\right)\left(5x+3\right)=30\)
\(\Leftrightarrow25x^2+10x+1-25x^2+9=30\)
\(\Leftrightarrow10x=20\)
hay x=2
b: Ta có: \(\left(x-1\right)\left(x^2+x+1\right)-x\left(x+2\right)\left(x-2\right)=5\)
\(\Leftrightarrow x^3-1-x^3+4x=5\)
\(\Leftrightarrow4x=6\)
hay \(x=\dfrac{3}{2}\)
27 tháng 9 2021
a) \(\Leftrightarrow2\left|3x-1\right|=\dfrac{4}{5}\)
\(\Leftrightarrow\left|3x-1\right|=\dfrac{2}{5}\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-1=\dfrac{2}{5}\\3x-1=-\dfrac{2}{5}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{15}\\x=\dfrac{1}{5}\end{matrix}\right.\)
b)TH1: \(x\ge3\)
\(\Leftrightarrow x+5+x-3=9\Leftrightarrow2x=7\Leftrightarrow x=\dfrac{7}{2}\left(tm\right)\)
TH2: \(-5\le x< 3\)
\(\Leftrightarrow x+5-x+3=9\Leftrightarrow8=9\left(VLý\right)\)
TH3: \(x< -5\)
\(\Leftrightarrow-x-5-x+3=9\Leftrightarrow2x=-11\Leftrightarrow x=-\dfrac{11}{2}\left(tm\right)\)
TH
27 tháng 9 2021
\(a,2.|3x-1|-\dfrac{3}{4}=\dfrac{1}{20}\)
\(2.|3x-1|=\dfrac{1}{20}+\dfrac{3}{4}\)
\(2.|3x-1|=\dfrac{4}{5}\)
\(|3x-1|=\dfrac{4}{5}:2\)
\(|3x-1|=\dfrac{2}{5}\)
\(\Rightarrow3x-1=\pm\dfrac{2}{5}\)
\(3x-1=\dfrac{2}{5}\)
\(3x=\dfrac{2}{5}+1\)
\(3x=\dfrac{7}{5}\)
\(x=\dfrac{7}{5}:3\)
\(x=\dfrac{7}{15}\)
\(3x-1=-\dfrac{2}{5}\)
\(3x=-\dfrac{2}{5}+1\)
\(3x=\dfrac{3}{5}\)
\(x=\dfrac{3}{5}:3\)
\(x=\dfrac{1}{5}\)
21 tháng 9 2021
\(c,\Rightarrow\left[{}\begin{matrix}-2\left(x+2\right)+\left(4-x\right)=11\left(x< -2\right)\\2\left(x+2\right)+\left(4-x\right)=11\left(-2\le x\le4\right)\\2\left(x+2\right)+\left(x-4\right)=11\left(x>4\right)\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{11}{3}\left(tm\right)\\x=3\left(tm\right)\\x=\dfrac{11}{3}\left(ktm\right)\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{11}{3}\end{matrix}\right.\)
21 tháng 9 2021
\(a,\Rightarrow\left[{}\begin{matrix}x+\dfrac{5}{2}=3x+1\\x+\dfrac{5}{2}=-3x-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=-\dfrac{7}{8}\end{matrix}\right.\)
23 tháng 11 2021
\(a,\Rightarrow\left[{}\begin{matrix}5x+1=\dfrac{6}{7}\\5x+1=-\dfrac{6}{7}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}5x=\dfrac{1}{7}\\5x=-\dfrac{13}{7}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{35}\\x=-\dfrac{13}{35}\end{matrix}\right.\\ b,\Rightarrow\left(-\dfrac{1}{8}\right)^x=\dfrac{1}{64}=\left(-\dfrac{1}{8}\right)^2\Rightarrow x=2\\ c,\Rightarrow\left(x-2\right)\left(2x+3\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=2\\x=-\dfrac{3}{2}\end{matrix}\right.\\ d,\Rightarrow\left(x+1\right)^{x+10}-\left(x+1\right)^{x+4}=0\\ \Rightarrow\left(x+1\right)^{x+4}\left[\left(x+1\right)^6-1\right]=0\\ \Rightarrow\left[{}\begin{matrix}x+1=0\\\left(x+1\right)^6=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x+1=0\\x+1=1\\x+1=-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-1\\x=0\\x=-2\end{matrix}\right.\\ e,\Rightarrow\dfrac{3}{4}\sqrt{x}=\dfrac{5}{6}\left(x\ge0\right)\\ \Rightarrow\sqrt{x}=\dfrac{10}{9}\Rightarrow x=\dfrac{100}{81}\)
4 tháng 12 2023
a: \(-\dfrac{3}{2}x+\dfrac{1}{4}=\dfrac{1}{2}\left(x+1\right)\)
=>\(-\dfrac{3}{2}x+\dfrac{1}{4}=\dfrac{1}{2}x+\dfrac{1}{2}\)
=>\(-\dfrac{3}{2}x-\dfrac{1}{2}x=\dfrac{1}{2}-\dfrac{1}{4}\)
=>\(-2x=\dfrac{1}{4}\)
=>\(2x=-\dfrac{1}{4}\)
=>\(x=-\dfrac{1}{4}:2=-\dfrac{1}{8}\)
b: ĐKXĐ: x>=0
\(\left(6-3\sqrt{x}\right)\left(\left|x\right|-7\right)=0\)
=>\(\left\{{}\begin{matrix}6-3\sqrt{x}=0\\\left|x\right|-7=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}3\sqrt{x}=6\\\left|x\right|=7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x}=2\\\left[{}\begin{matrix}x=7\left(nhận\right)\\x=-7\left(loại\right)\end{matrix}\right.\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=7\left(nhận\right)\\x=4\left(nhận\right)\end{matrix}\right.\)
HQ
Hà Quang Minh
Giáo viên
19 tháng 9 2023
a)
\(\begin{array}{l}x:{\left( {\frac{{ - 1}}{2}} \right)^3} = - \frac{1}{2}\\x = - \frac{1}{2}.{\left( {\frac{{ - 1}}{2}} \right)^3}\\x = {\left( {\frac{{ - 1}}{2}} \right)^4}\\x = \frac{1}{{16}}\end{array}\)
Vậy \(x = \frac{1}{{16}}\).
b)
\(\begin{array}{l}x.{\left( {\frac{3}{5}} \right)^7} = {\left( {\frac{3}{5}} \right)^9}\\x = {\left( {\frac{3}{5}} \right)^9}:{\left( {\frac{3}{5}} \right)^7}\\x = {\left( {\frac{3}{5}} \right)^2}\\x = \frac{9}{{25}}\end{array}\)
Vậy \(x = \frac{9}{{25}}\).
c)
\(\begin{array}{l}{\left( {\frac{{ - 2}}{3}} \right)^{11}}:x = {\left( {\frac{{ - 2}}{3}} \right)^9}\\x = {\left( {\frac{{ - 2}}{3}} \right)^{11}}:{\left( {\frac{{ - 2}}{3}} \right)^9}\\x = {\left( {\frac{{ - 2}}{3}} \right)^2}\\x = \frac{4}{9}.\end{array}\)
Vậy \(x = \frac{4}{9}\).
d)
\(\begin{array}{l}x.{\left( {0,25} \right)^6} = {\left( {\frac{1}{4}} \right)^8}\\x.{\left( {\frac{1}{4}} \right)^6} = {\left( {\frac{1}{4}} \right)^8}\\x = {\left( {\frac{1}{4}} \right)^8}:{\left( {\frac{1}{4}} \right)^6}\\x = {\left( {\frac{1}{4}} \right)^2}\\x = \frac{1}{{16}}\end{array}\)
Vậy \(x = \frac{1}{{16}}\).
11 tháng 11 2021
a) \(2x\left(x-5\right)-x\left(3+2x\right)=26\)
\(\Rightarrow2x^2-10x-3x-2x^2=26\)
\(\Rightarrow-13x=26\Rightarrow x=-2\)
b) \(3x\left(1-2x\right)+2\left(3x+7\right)=29\)
\(\Rightarrow3x-6x^2+6x+14=29\)
\(\Rightarrow-6x^2+9x-15=0\)
\(\Rightarrow-6\left(x^2-\dfrac{3}{2}x+\dfrac{9}{16}\right)-\dfrac{93}{8}=0\)
\(\Rightarrow-6\left(x-\dfrac{3}{4}\right)^2-\dfrac{93}{8}=0\)(vô lý)
Vậy \(S=\varnothing\)
6 tháng 12 2023
\(a,1-3\left|2x-3\right|=-\dfrac{1}{2}\\ 3\left|2x-3\right|=1+\dfrac{1}{2}\\ 3\left|2x-3\right|=\dfrac{3}{2}\\ \left|2x-3\right|=\dfrac{3}{2}:3\\ \left|2x-3\right|=\dfrac{9}{2}\\ \Rightarrow\left[{}\begin{matrix}2x-3=\dfrac{9}{2}\\2x-3=-\dfrac{9}{2}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=\dfrac{15}{2}\\2x=-\dfrac{3}{2}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{15}{4}\\x=-\dfrac{3}{4}\end{matrix}\right.\)
Vậy `x in {15/4;-3/4}`
\(b,\left(\left|x\right|-0,2\right)\left(x^3-8\right)=0\\ \left(\left|x\right|-0,2\right)\left(x-2\right)\left(x^2+2x+4\right)=0\\ \Rightarrow\left[{}\begin{matrix}\left|x\right|-0,2=0\\x-2=0\\x^2+2x+4=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}\left|x\right|=0,2\\x=2\\\left(x+1\right)^2+3=0\left(lọai\right)\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=0,2\\x=-0,2\\x=2\end{matrix}\right.\)
Vậy `x in {+-0,2;2}`
8 tháng 10 2023
`#3107.101107`
a)
`-2/3(x + 1) = 1/6 - x`
`=> -2/3x - 2/3 = 1/6 - x`
`=> -2/3x + x = 1/6 + 2/3`
`=> 1/3x = 5/6`
`=> x = 5/6 \div 1/3`
`=> x =5/2`
Vậy, `x = 5/2`
b)
`3(x + 1/3) - 1/2(x + 2) = 5/2x - 1`
`=> 3x + 1 - 1/2x - 1 = 5/2x - 1`
`=> 3x - 1/2x - 5/2x = -1`
`=> 0x = -1` (vô lý)
Vậy, `x` không có giá trị thỏa mãn.
8 tháng 10 2023
a: \(\Leftrightarrow-\dfrac{2}{3}x-\dfrac{2}{3}=\dfrac{1}{6}-x\)
=>\(\dfrac{1}{3}x=\dfrac{1}{6}+\dfrac{2}{3}=\dfrac{5}{6}\)
=>\(x=\dfrac{5}{6}\cdot3=\dfrac{5}{2}\)
b: \(\Leftrightarrow3x+1-\dfrac{1}{2}x-1=\dfrac{5}{2}x-1\)
=>\(\dfrac{5}{2}x=\dfrac{5}{2}x-1\)
=>0=-1(vô lý)
HQ
Hà Quang Minh
Giáo viên
9 tháng 10 2023
a)
\(\begin{array}{l}\left( {9x - {2^3}} \right):5 = 2\\9x - {2^3} = 2.5\\9x - 8 = 10\\9x = 18\\x = 2\end{array}\)
Vậy \(x = 2\)
b)
\(\begin{array}{l}\left[ {{3^4} - \left( {{8^2} + 14} \right):13} \right]x = {5^3} + {10^2}\\\left[ {81 - \left( {64 + 14} \right):13} \right]x = 125 + 100\\\left[ {81 - 78:13} \right]x = 125 + 100\\\left[ {81 - 6} \right]x = 225\\75x = 225\\x = 3\end{array}\)
Vậy \(x = 3\)
a, | x - 3 | - ( -3 ) = 4
= | x - 3 | + 3 = 4
|x - 3| = 4 - 3
|x - 3| = 1
=> x - 3 = 1 hoặc x - 3 = ( - 1 )
x = 1 + 3 = 4 x = ( -1 ) + 3 = 2
vậy x = 4 hoặc x = 2
b, x - (1 - x) = 5 + (-1+ x )
= x - 1 + x = 5 + -1 + x
= x -x +x = 5 -1 + 1
= x = 5
a)\(\left|x-3\right|-\left(-3\right)=4\)
\(\left|x-3\right|+3=4\)
\(\left|x-3\right|=4-3\)
\(\left|x-3\right|=1\)
\(x-3=1\) hoặc \(x-3=-1\)
Tự giải TIẾP NHA