Bài 1 :

TH1 : \(x< 0;\)ta có :

\(-x=2x-1\)

\(\Rightarrow2x-\left(-x\right)=1\)

\(3x=1\)

\(x=\frac{1}{3}\)( Không thỏa mãn \(x< 0\))

TH2 : \(x\ge0;\)ta có :

\(x=2x-1\)

\(\Rightarrow x=1\)(thỏa mãn )

Vậy x = 1.

Bài 2 :

\(3^{21}=3^{20}.3=9^{10}.3\)

\(2^{31}=2^{30}.2=8^{10}.2\)

\(9^{10}>8^{10};3>2\Rightarrow3^{21}>2^{31}.\)

Vậy ...

so sánh 3²¹ và 2³¹

Ta có : \(3^{21}=3.3^{20}=3.9^{10}\)

\(2^{31}=2.2^{30}=2.8^{10}\)

Vậy : \(2.8^{10}< 3.9^{10}\Leftrightarrow3^{21}>2^{31}\)

Bài 1 đề là jz

Bài 2: 

ta có: 2³⁴ > 2³⁰ = (2³)¹⁰ = 8¹⁰ > 3¹⁰

=> 2³⁴ > 3¹⁰

Bài 1 đề là jz???

Bài 2: 

ta có: 2³⁴ > 2³⁰ = (2³)¹⁰ = 8¹⁰ > 3¹⁰

=> 2³⁴ > 3¹⁰

.

a: |x|-|y|=60

=>6|y|-|y|=60

=>5|y|=60

=>|y|=12

=>y=12 hoặc y=-12

y=12

=>\(x=6y=6\cdot12=72\)

y=-12

=>\(x=6y=6\cdot\left(-12\right)=-72\)

b: |x|+|y|<2

mà |x|+|y|>=0 và x,y nguyên

nên |x|+|y|∈{0;1}

=>(|x|;|y|)∈{(0;1);(0;0);(1;0)}

=>(x;y)∈{(0;0);(0;1);(0;-1);(1;0);(-1;0)}

d: xy+5x-7y=35

=>x(y+5)-7y-35=0

=>(x-7)(y+5)=0

=>x=7 và y=-5

e: xy+2x-3y=9

=>x(y+2)-3y-6=3

=>(x-3)(y+2)=3

=>(x-3;y+2)∈{(1;3);(3;1);(-1;-3);(-3;-1)}

=>(x;y)∈{(4;1);(6;-1);(2;-5);(0;-3)}

f: xy-2x+5y-12=0

=>x(y-2)+5y-10=2

=>(x+5)(y-2)=2

=>(x+5;y-2)∈{(1;2);(2;1);(-1;-2);(-2;-1)}

=>(x;y)∈{(-4;4);(-3;3);(-6;0);(-7;1)}

giải hộ e vs akk đang cần gấp

\(1,\\ \left(x-7\right)^{x+1}-\left(x-7\right)^{x+11}=0\\ \Leftrightarrow\left(x-7\right)^{x+1}\left[1-\left(x-7\right)^{10}\right]=0\\ \Leftrightarrow\left[{}\begin{matrix}\left(x-7\right)^{x+1}=0\\\left(x-7\right)^{10}=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-7=0\\x-7=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=7\\x=8\end{matrix}\right.\)

\(2,\\ a,\left|2x-3\right|>5\Leftrightarrow\left[{}\begin{matrix}2x-3< -5\\2x-3>5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x< -1\\x>4\end{matrix}\right.\\ b,\left|3x-1\right|\le7\Leftrightarrow\left[{}\begin{matrix}3x-1\le7\\1-3x\le7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x\le\dfrac{8}{3}\\x\ge-2\end{matrix}\right.\\ c,\cdot x< -\dfrac{3}{2}\\ \Leftrightarrow5-3x+\left(-2x-3\right)=7\Leftrightarrow2-5x=7\Leftrightarrow x=-1\left(ktm\right)\\ \cdot-\dfrac{3}{2}\le x\le\dfrac{5}{3}\\ \Leftrightarrow\left(5-3x\right)+\left(2x+3\right)=7\Leftrightarrow8-x=7\Leftrightarrow x=1\left(tm\right)\\ \cdot x>\dfrac{5}{3}\\ \Leftrightarrow\left(3x-5\right)+\left(2x+3\right)=7\Leftrightarrow5x-2=7\Leftrightarrow x=\dfrac{9}{5}\left(tm\right)\\ \Leftrightarrow S=\left\{1;\dfrac{9}{5}\right\}\)

Mai lam