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LT
6
22 tháng 7 2020
5/2+x=-3/5
x=-3/5-5/2
x=-31/10
vậy x=-31/10
-2x+5=23
-2x=23-5
-2x=18
x=18:-2
x=-9
vậy x=-9
B
22 tháng 7 2020
Bài làm :
\(\frac{5}{2}+x=-3,5\)
\(\Leftrightarrow x=-3,5-\frac{5}{2}=-6\)
\(-2x+5=23\)
\(\Leftrightarrow-2x=18\)
\(\Leftrightarrow x=-9\)
\(\frac{x}{21}=\frac{2}{7}-\frac{4}{3}\)
\(\Leftrightarrow\frac{x}{21}=-\frac{22}{21}\)
\(\Leftrightarrow x=-22\)
\(\frac{5}{3}x-\frac{2}{5}x=\frac{19}{10}\)
\(\Leftrightarrow x\left(\frac{5}{3}-\frac{2}{5}\right)=\frac{19}{10}\)
\(\Leftrightarrow\frac{19}{15}x=\frac{19}{10}\)
\(\Leftrightarrow x=\frac{3}{2}\)
Chúc bạn học tốt !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
21 tháng 8 2020
a) \(\left(\frac{7}{8}-\frac{3}{4}\right)\cdot\frac{1}{3}-\frac{2}{7}\cdot\left(3,5\right)=\left(\frac{7}{8}-\frac{3}{4}\right)\cdot\frac{1}{3}-\frac{2}{7}\cdot\frac{7}{2}\)
\(=\left(\frac{7}{8}-\frac{6}{8}\right)\cdot\frac{1}{3}-1=\frac{1}{8}\cdot\frac{1}{3}-1=\frac{1}{24}-\frac{24}{24}=-\frac{23}{24}\)
b) \(\left(\frac{3}{5}+0,415-\frac{3}{200}\right)\cdot2\frac{2}{3}\cdot0,25\)
\(=\left(\frac{3}{5}+\frac{83}{200}-\frac{3}{200}\right)\cdot\frac{8}{3}\cdot\frac{1}{4}\)
\(=\left(\frac{3}{5}+\frac{80}{200}\right)\cdot\frac{8}{3}\cdot\frac{1}{4}=\left(\frac{3}{5}+\frac{2}{5}\right)\cdot\frac{8}{3}\cdot\frac{1}{4}=1\cdot\frac{8}{3}\cdot\frac{1}{4}=1\cdot\frac{2}{3}\cdot\frac{1}{1}=\frac{2}{3}\)
c) \(\frac{5}{16}:0,125-\left(2\frac{1}{4}-0,6\right)\cdot\frac{10}{11}\)
\(=\frac{5}{16}:\frac{1}{8}-\left(\frac{9}{4}-\frac{3}{5}\right)\cdot\frac{10}{11}\)
\(=\frac{5}{16}\cdot8-\frac{33}{20}\cdot\frac{10}{11}=\frac{5}{2}-\frac{3}{2}=1\)
d) \(0,25:\left(10,3-9,8\right)-\frac{3}{4}=\frac{1}{4}:\left(\frac{103}{10}-\frac{98}{10}\right)-\frac{3}{4}\)
\(=\frac{1}{4}:\frac{1}{2}-\frac{3}{4}=\frac{1}{4}\cdot2-\frac{3}{4}=\frac{2}{4}-\frac{3}{4}=-\frac{1}{4}\)
Câu cuối tương tự
22 tháng 10 2019
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 \(x\in\left\{15;18;12\right\}\)
22 tháng 10 2019
\(1.6x^3-8=40\\ \Leftrightarrow6x^3=48\\ \Leftrightarrow x^3=8\Leftrightarrow x^3=2^3=\left(-2\right)^3\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)
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.
DN
0
10 tháng 4 2017
Câu 1:
(\(\frac{3x}{7}\)\(+\)\(1\))\(\div\)(\(-4\)) \(=-\)\(\frac{1}{28}\)
\(\Leftrightarrow\)\(\frac{3x}{7}\)\(+\)\(1=-\)\(\frac{1}{28}\)\(\times\)(\(-4\))
\(\Leftrightarrow\)\(\frac{3x}{7}\)\(+\)\(1=\)\(\frac{1}{7}\)
\(\Leftrightarrow\)\(\frac{3x}{7}\)\(=\)\(\frac{1}{7}\)\(-1\)
\(\Leftrightarrow\)\(\frac{3x}{7}\)\(=-\)\(\frac{6}{7}\)
\(\Leftrightarrow\)\(x\)\(=-\)\(\frac{6}{7}\)\(\div\)\(\frac{3}{7}\)
\(\Leftrightarrow\)\(x\)\(=\)\(-2\)
8 tháng 9 2016
a) \(\left|2x\right|-\left|-2,5\right|=\left|-7,5\right|\)
\(\Leftrightarrow\left|2x\right|-2,5=7,5\)
\(\Leftrightarrow\left|2x\right|=10\)
\(\Leftrightarrow\begin{cases}x\ge0\\2x=10\end{cases}\) hoặc \(\begin{cases}x< 0\\2x=-10\end{cases}\)
\(\Leftrightarrow\begin{cases}x\ge0\\x=5\left(tm\right)\end{cases}\) hoặc \(\begin{cases}x< 0\\x=-5\left(tm\right)\end{cases}\)
Vậy x={5;-5}
b)\(\left|3x\right|\cdot\left|-3,5\right|=\left|-2,8\right|\)
\(\Leftrightarrow\left|3x\right|\cdot3,5=2,8\)
\(\Leftrightarrow\left|3x\right|=\frac{4}{5}\)
\(\Leftrightarrow\begin{cases}x\ge0\\3x=\frac{4}{5}\end{cases}\) hoặc \(\begin{cases}x< 0\\3x=-\frac{4}{5}\end{cases}\)
\(\Leftrightarrow\begin{cases}x\ge0\\x=\frac{4}{15}\end{cases}\) hoặc \(\begin{cases}x< 0\\x=-\frac{4}{15}\end{cases}\)
Vậy x={4/15;-4/15}
c) \(\left(3x-5\right)\left(\frac{3}{2}x+2\right)\left(0,5x-10\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}3x-5=0\\\frac{3}{2}x+2=0\\0,5x-10=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=\frac{5}{3}\\x=-\frac{4}{3}\\x=20\end{array}\right.\)
EC
8 tháng 9 2016
a)|2x|-|-2,5|=|-7,5|
|2x|-2,5=7,5
|2x|=10
\(\Rightarrow\left[\begin{array}{nghiempt}2x=10\\2x=-10\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}x=5\\x=-5\end{array}\right.\)
Vậy x=5;-5
13 tháng 1 2019
1) 4x-(2x-5)=21
4x-2x+5=21
2x+5=21
2x=21 -5
2x=16
x=16/2
x=8
PA
7 tháng 1 2019
1) 14x-8x=10+5
x(14-8)=15
x6=15
x=15/6
2)5x-3x=30-15
2x=15
x=15/2
3)làm tương tự
VC
Vũ Cao Minh⁀ᶦᵈᵒᶫ ( ✎﹏IDΣΛ亗 )
CTV
VIP
26 tháng 8 2021
a) \(3x-10=2x+13\Leftrightarrow x=13+10=23\)
b) \(x+12=-5-x\Leftrightarrow2x=-17\Leftrightarrow x=-\frac{17}{2}\)
c) \(x+5=10-x\Leftrightarrow2x=5\Leftrightarrow x=\frac{5}{2}\)
d) \(6x+23=2x-12\Leftrightarrow4x=-35\Leftrightarrow x=-\frac{35}{4}\)
e) \(12-x=x+1\Leftrightarrow2x=11\Leftrightarrow x=\frac{11}{2}\)
f) \(14+4x=3x+20\Leftrightarrow x=20-14=6\)
VC
Vũ Cao Minh⁀ᶦᵈᵒᶫ ( ✎﹏IDΣΛ亗 )
CTV
VIP
26 tháng 8 2021
g) \(2\left(x-1\right)+3\left(x-2\right)=x-4\Leftrightarrow2x-2+3x-6=x-4\)
\(\Leftrightarrow5x-8=x-4\Leftrightarrow4x=4\Leftrightarrow x=1\)
h ) \(3\left(4-x\right)-2\left(x-1\right)=x+20\Leftrightarrow12-3x-2x+2=x+20\)
\(\Leftrightarrow14-5x=x+20\Leftrightarrow6x=-6\Leftrightarrow x=-1\)
i ) \(4\left(2x+7\right)-3\left(3x-2\right)=24\Leftrightarrow8x+28-9x+6=24\)
\(\Leftrightarrow34-x=24\Leftrightarrow x=34-24=10\)
k) \(3\left(x-2\right)+2x=10\Leftrightarrow3x-6+2x=5x-6=10\)
\(\Leftrightarrow5x=10+6=16\Leftrightarrow x=\frac{16}{5}\)
Tự KL cho các phần
2 và 2/3x 3,5-(1/10+20%):3/5
2/3x 3,5-(1/10+20%):3/5
= 2/3 x 35/10 - ( 1/10 + 20/100 ) : 3/5
= 2/3 x 7/2 - ( 1/10 - 1/5) : 3/5
= 7/3 - ( -1/10) : 3/5
= 73/30 : 3/5
= 73/18 \(\approx\) 4.05
=> 2 < 4,05
\(\dfrac{2}{3}\times3,5-\left(\dfrac{1}{10}+20\%\right):\dfrac{3}{5}\\ =\dfrac{2}{3}\times\dfrac{7}{2}-\left(\dfrac{1}{10}+\dfrac{20}{100}\right).\dfrac{5}{3}\\ =\dfrac{7}{3}-\left(\dfrac{1}{10}+\dfrac{2}{10}\right).\dfrac{5}{3}\\ =\dfrac{7}{3}-\dfrac{3}{10}.\dfrac{5}{3}\\ =\dfrac{7}{3}-\dfrac{5}{10}\\ =\dfrac{7}{3}-\dfrac{1}{2}\\ =\dfrac{14}{6}-\dfrac{3}{6}=\dfrac{11}{6}\)
\(2=\dfrac{12}{6}>\dfrac{11}{6}\)
Vậy : \(2>\dfrac{2}{3}x3,5-\left(\dfrac{1}{10}+20\%\right):\dfrac{3}{5}\)