a) x - 1/2 = 3/5

x = 3/5 + 1/2

x = 11/10

b) x - 1/2 = -2/3

x = -2/3 + 1/2

x = -1/6

c) 2/5 - x = 0,25

x = 2/5 - 0,25

x = 2/5 - 1/4

x = 3/20

12:

a: ĐKXĐ: -2x+3>=0

=>x<=3/2

b: ĐKXĐ: x^2<>0

=>x<>0

c: ĐKXĐ: x+3>0

=>x>-3

d:ĐKXĐ: -5/x^2+6>=0

=>x^2+6<0

=>x thuộc rỗng

4. 

a) Ta có: \(\sqrt{x}=3\)

\(\Leftrightarrow x=3^2=9\)

 Vậy \(x=9\)

b) Ta có: \(\sqrt{x}=\sqrt{5}\)

\(\Leftrightarrow x=\sqrt{5}^2=5\)

Vậy \(x=5\)

c) Ta có: \(\sqrt{x}=0\)

\(\Leftrightarrow x=0^2=0\)

Vậy \(x=0\)

d) Ta có: \(\sqrt{x}=-2\)

\(\Leftrightarrow x=-2^2=4\)

Vậy \(x=4\)

x thuộc 6,51; 6,52; 6,53;...

x = 6,51 ; 6,52 ; 6,53 ; 6,54 ; 6,55 ; 6,56 ; 6,57 ; 6,58 ; 6;59

x=6,51;6,52;6,53;6,54;6,55;6,56;6,57;6,58;6,59

\(\left(x+1\right)^3-\left(x-1\right)\left(x^2+x+1\right)-2=0\)

\(\Rightarrow x^3+3x^2+3x+1-x^3+1-2=0\)

\(\Rightarrow3x^2+3x=0\Rightarrow3x\left(x+1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)

\(=x^3+3x^2+3x+1-x^3+1-2=0\\ \Leftrightarrow3x^2+3x=0\\ \Leftrightarrow3x\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)

Ta có :

\(\frac{x-3}{97}+\frac{x-27}{73}+\frac{x-67}{33}+\frac{x-73}{27}=4\)

\(\Leftrightarrow\left(\frac{x-3}{97}-1\right)+\left(\frac{x-27}{73}-1\right)+\left(\frac{x-67}{33}-1\right)+\left(\frac{x-73}{27}-1\right)=0\)

\(\Leftrightarrow\frac{x-100}{97}+\frac{x-100}{73}+\frac{x-100}{33}+\frac{x-100}{27}=0\)

\(\Leftrightarrow\left(x-100\right)\left(\frac{1}{97}+\frac{1}{73}+\frac{1}{33}+\frac{1}{27}\right)=0\)

Vì \(\frac{1}{97}+\frac{1}{73}+\frac{1}{33}+\frac{1}{27}>0\) Nên \(x-100=0\)

\(\Leftrightarrow x=100\)

Vậy \(x=100\)

\(\Leftrightarrow\frac{x-3}{87}+\frac{x-27}{79}+\frac{x-67}{33}+\frac{x-73}{27}-4=0\)

\(\Leftrightarrow\left(\frac{x-3}{97}-1\right)+\left(\frac{x-27}{73}-1\right)+\left(\frac{x-67}{33}-1\right)+\left(\frac{x-73}{27}-1\right)=0\)

\(\Leftrightarrow\left(\frac{x-3-97}{97}\right)+\left(\frac{x-27-73}{73}\right)+\left(\frac{x-67-33}{33}\right)+\left(\frac{x-73-27}{27}\right)=0\)

\(\Leftrightarrow\frac{x-100}{97}+\frac{x-100}{73}+\frac{x-100}{33}+\frac{x-100}{27}=0\)

\(\Leftrightarrow\left(x-100\right)\left(\frac{1}{97}+\frac{1}{73}+\frac{1}{33}+\frac{1}{27}\right)=0\)

Vì \(\frac{1}{97}+\frac{1}{73}+\frac{1}{33}+\frac{1}{27}\ne0\)

\(\Rightarrow x-100=0\Leftrightarrow x=100\)

\(\frac{x-3}{97}+\frac{x-27}{73}+\frac{x-67}{33}+\frac{x-73}{27}=4\)    

\(\frac{x-3}{97}-1+\frac{x-27}{73}-1+\frac{x-67}{33}-1+\frac{x-73}{27}-1=4-1-1-1-1\)     

\(\frac{x-100}{97}+\frac{x-100}{73}+\frac{x-100}{33}+\frac{x-100}{27}=0\)    

\(\left(x-100\right)\left(\frac{1}{97}+\frac{1}{73}+\frac{1}{33}+\frac{1}{27}\right)=0\)   

\(x-100=0\) 

\(x=100\)

- Bài này phải có điều kiện \(x>0\) thì mới làm được nhé bạn.

\(P=\dfrac{x^3+2021}{x}=x^2+\dfrac{2021}{x}=x^2+\dfrac{2021}{2x}+\dfrac{2021}{2x}\)

Áp dụng bất đẳng thức Cauchy cho 3 số dương ta có:

\(x^2+\dfrac{2021}{2x}+\dfrac{2021}{2x}\ge3\sqrt[3]{x^2.\dfrac{2021}{2x}.\dfrac{2021}{2x}}=3\sqrt[3]{\dfrac{2021^2}{4}}\)

Dấu "=" xảy ra khi \(x^2=\dfrac{2021}{2x}\Leftrightarrow x=\sqrt[3]{\dfrac{2021}{2}}\)

Vậy \(MinP=3\sqrt[3]{\dfrac{2021^2}{4}}\)

\(b,N=\left(2x-1\right)^2-4\ge-4\\ N_{min}=-4\Leftrightarrow x=\dfrac{1}{2}\\ c,P=\left(2x-5\right)^2+6\left(2x-5\right)+9-4\\ P=\left(2x-5+3\right)^2-4=\left(2x-2\right)^2-4\ge-4\\ P_{min}=-4\Leftrightarrow x=1\\ d,Q=\left(x^2-2x+1\right)+\left(y^2+4y+4\right)+1\\ Q=\left(x-1\right)^2+\left(y+2\right)^2+1\ge1\\ Q_{min}=1\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)

6a.

$M=x^2-x+1=(x^2-x+\frac{1}{4})+\frac{3}{4}$

$=(x-\frac{1}{2})^2+\frac{3}{4}\geq \frac{3}{4}$

Vậy $M_{\min}=\frac{3}{4}$ khi $x-\frac{1}{2}=0\Leftrightarrow x=\frac{1}{2}$

a: x/2=-5/y

=>xy=-10

=>\(\left(x,y\right)\in\left\{\left(1;-10\right);\left(-10;1\right);\left(-1;10\right);\left(10;-1\right);\left(2;-5\right);\left(-5;2\right);\left(-2;5\right);\left(5;-2\right)\right\}\)

b: =>xy=12

mà x>y>0

nên \(\left(x,y\right)\in\left\{\left(12;1\right);\left(6;2\right);\left(4;3\right)\right\}\)

c: =>(x-1)(y+1)=3

=>\(\left(x-1;y+1\right)\in\left\{\left(1;3\right);\left(3;1\right);\left(-1;-3\right);\left(-3;-1\right)\right\}\)

=>\(\left(x,y\right)\in\left\{\left(2;2\right);\left(4;0\right);\left(0;-4\right);\left(-2;-2\right)\right\}\)

d: =>y(x+2)=5

=>\(\left(x+2;y\right)\in\left\{\left(1;5\right);\left(5;1\right);\left(-1;-5\right);\left(-5;-1\right)\right\}\)

=>\(\left(x,y\right)\in\left\{\left(-1;5\right);\left(3;1\right);\left(-3;-5\right);\left(-7;-1\right)\right\}\)