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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

1) (3x + 9)(3x - 6) = 0
=> \(\orbr{\begin{cases}3x+9=0\\3x-6=0\end{cases}}\)
=> \(\orbr{\begin{cases}3x=-9\\3x=6\end{cases}}\)
=> \(\orbr{\begin{cases}x=-3\\x=2\end{cases}}\)
Vậy ...
b) (2x + 15) - 25 = 47 - (10 - x)
=> 2x - 10 = 37 + x
=> 2x - x = 37 + 10
=> x = 47
3, tương tự
4) |4 - 3x| = 8
=> \(\orbr{\begin{cases}4-3x=8\\4-3x=-8\end{cases}}\)
=> \(\orbr{\begin{cases}3x=-4\\3x=12\end{cases}}\)
=> \(\orbr{\begin{cases}x=-\frac{4}{3}\\x=4\end{cases}}\)
Vì x là số nguyên nên ...
còn lại tương tự

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.


a) -3x+4+5x=-10-x
-3x+4+5x+10+x=0
(-3x+5x+x)+10=0
3x+10=0
3x=-10
x=\(\dfrac{-10}{3}\)
Vậy x=\(\dfrac{-10}{3}\)
b)-x+1=-3x-8
-x+1+3x+8=0
(-x+3x)+(1+8)=0
2x+9=0
2x=-9
x=\(\dfrac{-9}{2}\)
Vậy x=\(\dfrac{-9}{2}\)
c)8-(x-1)=10+(x+5)
8-x+1=10+x+5
9-x=15+x
9-x-15-x=0
(9-15)-(x+x)=0
-6-2x=0
2x=-6
x=-3
Vậy x=-3
d)100+(x+7)-(-2x+3)=8+(x+100)
100+x+7+2x-3=8+x+100
(x+2x)+(100+7-3)=(8+100)+x
3x+104=108+x
3x+104-108-x=0
(3x-x)+(104-108)=0
2x-4=0
2x=4
x=2
Vậy x=2
e, \(\left|2x+5\right|=\left|x-1\right|\)
\(\Rightarrow\left\{{}\begin{matrix}2x+5=1-x\\2x+5=x-1\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}3x=-4\\x=-6\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{4}{3}\\x=-6\end{matrix}\right.\)
g, \(\left|-x+4\right|=\left|-3x-8\right|\)
\(\Rightarrow\left\{{}\begin{matrix}-x+4=3x+8\\-x+4=-3x-8\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}-4x=4\\2x=-12\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-1\\x=-6\end{matrix}\right.\)
h, \(\left|x+4\right|=\left|-3-8\right|\)
\(\Rightarrow\left|x+4\right|=\left|-11\right|=11\)
\(\Rightarrow\left\{{}\begin{matrix}x+4=-11\\x+4=11\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-15\\x=7\end{matrix}\right.\)
Chúc bạn học tốt!!!
a) Ta có: \(\left|x\right|-\left(-9\right)^3=-100\)
\(\Leftrightarrow\left|x\right|-\left(-729\right)=-100\)
\(\Leftrightarrow\left|x\right|+729=-100\)
\(\Leftrightarrow\left|x\right|=-100-729=-829\)(vô lý)
Vậy: \(x\in\varnothing\)
b) Ta có: \(3x-\left(-10\right)=2x+50\)
\(\Leftrightarrow3x+10-2x-50=0\)
\(\Leftrightarrow x-40=0\)
\(\Leftrightarrow x=40\)
Vậy: x=40