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a)
\(\frac{2}{3}+\frac{1}{3}:x=\frac{3}{5}\)
\(\frac{1}{3}:x=\frac{3}{5}-\frac{2}{3}\)
\(\frac{1}{3}:x=\frac{-1}{15}\)
\(x=\frac{1}{3}:\frac{-1}{15}\)
\(x=-5\)
b)
\(\frac{10}{3x}+\frac{67}{4}=-13.25\)
\(\frac{10}{3x}+\frac{67}{4}=\frac{-53}{4}\)
\(\frac{10}{3x}=\frac{-53}{4}+\frac{67}{4}\)
\(\frac{10}{3x}=\frac{7}{2}\)
\(\Rightarrow10.2=7.3x\)
\(20=21x\)
\(x=\frac{20}{21}\)
c)
x+30%x=-1.3
x+0.3x=-1.3
x (1+0.3) = -1.3
x . 1.3 = -1.3
x = -1.3 : 1.3
x = 1
d)
\(\left(\frac{14}{5x}-50\right):\frac{2}{3}=51\)
\(\frac{14}{5x}-50=51.\frac{2}{3}\)
\(\frac{14}{5x}-50=34\)
\(\frac{14}{5x}=34+50\)
\(\frac{14}{5x}=84\)
\(84.5x=14\)
420x = 14
\(x=\frac{1}{30}\)
a, \(\frac{2}{3}+\frac{1}{3}:x=\frac{3}{5}\)
=> \(\frac{1}{3}:x=\frac{3}{5}-\frac{2}{3}\)
=>\(\frac{1}{3}:x=\frac{-1}{15}\)
=> \(x=\frac{1}{3}:\frac{-1}{15}\)
=> \(x=\frac{-1}{5}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
c)
\(4\left(3x-4\right)-2=18\)
<=> \(12x-16-2=18\)
<=> \(12x=36\)
<=> \(x=3\)
Vậy x=3
d)
\(\left(3x-10\right):10=50\)
<=> \(3x-10=500\)
<=> \(3x=510\)
<=> x= \(170\)
Vậy x= 170
f)
\(x-\left[42+\left(-25\right)\right]=-8\)
<=> \(x-17=-8\)
<=> x= \(9\)
Vậy x=9
h)
\(x+5=20-\left(12-7\right)\)
<=> \(x+5=15\)
<=> \(x=10\)
Vậy x= 10
k)
\(\left|x-5\right|=7-\left(-3\right)\)
<=> \(\left|x-5\right|=10\)
* Với \(x>=5\) ; ta được:
\(x-5=10\)
<=> x= 15 (thoả mãn điều kiện )
*Với \(x< 5\) ; ta được:
\(-\left(x-5\right)=10\)
<=> \(-x+5=10\)
<=> \(-x=5\)
<=> \(x=-5\) (thoả mãn điều kiện)
Vậy x=15 ; x= -5
i)
\(\left|x-5\right|=\left|7\right|\)
<=> \(\left|x-5\right|=7\)
*Với \(x>=5\) ; ta được:
\(x-5=7\)
<=> \(x=12\) (thoả mãn)
*Với \(x< 5\) ; ta được:
\(-\left(x-5\right)=7\)
<=> \(-x=2\)
<=> \(x=-2\) (thoả mãn)
Vậy x= 12; x= -2
m)
\(2^{x+1}.2^{2009}=2^{2010}\)
<=> \(2^{x+1+2009}=2^{2010}\)
<=> \(2^{x+2010}=2^{2010}\)
=> \(x+2010=2010\)
=> \(x=0\)
Vậy x=0
n)
\(10-2x=25-3x\)
<=>\(x=15\)
Vậy x=15
![](https://rs.olm.vn/images/avt/0.png?1311)
Trả lời:
A = ( 2x - 7 )4
Ta có: \(\left(2x-7\right)^4\ge0\forall x\)
Dấu "=" xảy ra khi 2x - 7 = 0 <=> 2x = 7 <=> x = 7/2
Vậy GTNN của A = 0 khi x = 7/2
B = ( x + 1 )10 + ( y - 2 )20 + 7
Ta có: \(\left(x+1\right)^{10}\ge0\forall x;\left(y-2\right)^{20}\ge0\forall y\)
\(\Leftrightarrow\left(x+1\right)^{10}+\left(y-2\right)^{20}\ge0\forall x;y\)
\(\Leftrightarrow\left(x+1\right)^{10}+\left(y-2\right)^{20}+7\ge7\forall x;y\)
Dấu "=" xảy ra khi x + 1 = 0 <=> x = -1 và y - 2 = 0 <=> y = 2
Vậy GTNN của B = 7 khi x = -1 và y = 2
C = ( 3x - 4 )100 + ( 5y + 1 )50 - 20
Ta có: \(\left(3x-4\right)^{100}\ge0\forall x;\left(5y+1\right)^{50}\ge0\forall y\)
\(\Leftrightarrow\left(3x-4\right)^{100}+\left(5y+1\right)^{50}\ge0\forall x;y\)
\(\Leftrightarrow\left(3x-4\right)^{100}+\left(5y+1\right)^{50}-20\ge-20\forall x;y\)
Dấu "=" xảy ra khi 3x - 4 = 0 <=> x = 4/3 và 5y + 1 = 0 <=> y = -1/5
Vậy GTNN của C = -20 khi x = 4/3 và y = -1/5
D = ( 2x + 3 )20 + ( 3y - 4 )10 + 1000
Ta có: \(\left(2x+3\right)^{20}\ge0\forall x;\left(3y-4\right)^{10}\ge0\forall y\)
\(\Leftrightarrow\left(2x+3\right)^{20}+\left(3y-4\right)^{10}\ge0\forall x;y\)
\(\Leftrightarrow\left(2x+3\right)^{20}+\left(3y-4\right)^{10}+100^0\ge1\forall x;y\)
Dấu "=" xảy ra khi 2x + 3 = 0 <=> x = -3/2 và 3y - 4 = 0 <=> y = 4/3
Vậy GTNN của D = 1 khi x = -3/2 và y = 4/3
E = ( x - y )50 + ( y - 2 )60 + 3
Ta có: \(\left(x-y\right)^{50}\ge0\forall x;y\); \(\left(y-2\right)^{60}\ge0\forall y\)
\(\Leftrightarrow\left(x-y\right)^{50}+\left(y-2\right)^{60}\ge0\forall x;y\)
\(\Leftrightarrow\left(x-y\right)^{50}+\left(y-2\right)^{60}+3\ge3\forall x;y\)
Dấu "=" xảy ra khi x - y = 0 <=> x = y và y - 2 = 0 <=> y = 2
Vậy GTNN của E = 3 khi x = y = 2
![](https://rs.olm.vn/images/avt/0.png?1311)
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.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\dfrac{2}{3}x-\dfrac{1}{2}=\dfrac{1}{10}\)
\(\dfrac{2}{3}x=\dfrac{1}{10}+\dfrac{1}{2}=\dfrac{3}{5}\)
\(x=\dfrac{3}{5}:\dfrac{2}{3}=\dfrac{9}{10}\)
b) \(\dfrac{39}{7}:x=13\)
\(x=\dfrac{\dfrac{39}{7}}{13}=\dfrac{3}{7}\)
c) \(\left(\dfrac{14}{5}x-50\right):\dfrac{2}{3}=51\)
\(\dfrac{14}{5}x-50=51\cdot\dfrac{2}{3}=34\)
\(\dfrac{14}{5}x=34+50=84\)
\(x=\dfrac{84}{\dfrac{14}{5}}=30\)
d) \(\left(x+\dfrac{1}{2}\right)\left(\dfrac{2}{3}-2x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=0\\\dfrac{2}{3}-2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=\dfrac{1}{3}\end{matrix}\right.\)
e) \(\dfrac{2}{3}x-\dfrac{1}{2}x=\dfrac{5}{12}\)
\(\dfrac{1}{6}x=\dfrac{5}{12}\)
\(x=\dfrac{5}{12}:\dfrac{1}{6}=\dfrac{5}{2}\)
g) \(\left(x\cdot\dfrac{44}{7}+\dfrac{3}{7}\right)\dfrac{11}{5}-\dfrac{3}{7}=-2\)
\(\left(x\cdot\dfrac{44}{7}+\dfrac{3}{7}\right)\cdot\dfrac{11}{5}=-2+\dfrac{3}{7}=-\dfrac{11}{7}\)
\(x\cdot\dfrac{44}{7}+\dfrac{3}{7}=-\dfrac{11}{7}:\dfrac{11}{5}=-\dfrac{5}{7}\)
\(\dfrac{44}{7}x=-\dfrac{5}{7}-\dfrac{3}{7}=-\dfrac{8}{7}\)
\(x=-\dfrac{8}{7}:\dfrac{44}{7}=-\dfrac{2}{11}\)
h) \(\dfrac{13}{4}x+\left(-\dfrac{7}{6}\right)x-\dfrac{5}{3}=\dfrac{5}{12}\)
\(\dfrac{25}{12}x-\dfrac{5}{3}=\dfrac{5}{12}\)
\(\dfrac{25}{12}x=\dfrac{5}{12}+\dfrac{5}{3}=\dfrac{25}{12}\)
\(x=1\)
Mỏi tay woa bn làm nốt nha!!
![](https://rs.olm.vn/images/avt/0.png?1311)
c: =>2/3x=1/10+1/2=1/10+5/10=6/10=3/5
hay \(x=\dfrac{3}{5}:\dfrac{2}{3}=\dfrac{9}{10}\)
d: \(\Leftrightarrow\dfrac{4}{9}:x=\dfrac{2}{3}-\dfrac{3}{5}=\dfrac{1}{15}\)
hay \(x=\dfrac{4}{9}:\dfrac{1}{15}=\dfrac{4}{9}\cdot15=\dfrac{20}{3}\)
f: (x+1/2)(2/3-2x)=0
=>x+1/2=0 hoặc 2/3-2x=0
=>x=-1/2 hoặc x=1/3
LÀM LUN NHA
2/3x-1/2+x=1/10
5/3x=1/10+1/2
5/3x=3/5
X=9/25
**** CHO MK NHA