\(\left(3x-3\right)^2+\left(4y+2\right)^2=0\)

Ta có:

`(3x-3)^2>=0` với mọi x

`(4y+2)^2>=0` với mọi x

`=>(3x-3)^2+(4y+2)^2>=0` với mọi x,y

Mặt khác: `(3x-3)^2+(4y+2)^2=0`

Dấu "=" xảy ra: `3x-3=0` và `4y+2=0`

`=>3x=3` và `4y=-2`

`=>x=3/3=1` và `y=-2/4=-1/2` 

\(A=10\cdot\dfrac{4}{7}-1,38-7,62\\ =10\cdot\dfrac{4}{7}-\left(1,38+7,62\right)\\ =\dfrac{10\cdot4}{7}-9\\ =\dfrac{40}{7}-\dfrac{63}{7}\\ =\dfrac{40-63}{7}\\ =\dfrac{-23}{7}\)

Hai góc phụ nhau nên \(\widehat{M}+\widehat{N}=90^0\)

mà \(\widehat{M}-\widehat{N}=20^0\)

nên \(\widehat{M}=\dfrac{90^0+20^0}{2}=55^0;\widehat{N}=55^0-20^0=35^0\)

15x⁴ + 7x⁴ + (−20x²)²

= 22x⁴ + 400x⁴

= 422x⁴

Tại x = -1, ta có:

422x⁴ = 422.(−1)⁴  422

\(\left(2x-1\right)\left(3x+1\right)+\left(3x+4\right)\left(3x-2\right)=5\)

=>\(6x^2+2x-3x-1+9x^2-6x+12x-8=5\)

=>\(15x^2+5x-9-5=0\)

=>\(15x^2+5x-14=0\)

\(\Delta=5^2-4\cdot15\cdot\left(-14\right)=25+60\cdot14=25+840=865>0\)

Do đó: Phương trình có hai nghiệm phân biệt là:

\(\left[{}\begin{matrix}x=\dfrac{-5-\sqrt{865}}{2\cdot15}=\dfrac{-5-\sqrt{865}}{30}\\x=\dfrac{-5+\sqrt{865}}{30}\end{matrix}\right.\)

a; - \(\dfrac{3}{5}\) + \(\dfrac{7}{5}\) = \(\dfrac{-3+7}{5}\) = \(\dfrac{4}{5}\)

b; \(\dfrac{3}{5}\) - \(\dfrac{-9}{5}\) = \(\dfrac{3-\left(-9\right)}{5}\) = \(\dfrac{3+9}{5}\) = \(\dfrac{12}{5}\)

c; \(\dfrac{23}{-11}\) - \(\dfrac{-3}{11}\) = \(\dfrac{-23}{11}\) + \(\dfrac{3}{11}\) = \(\dfrac{-20}{11}\)

d; -2\(\dfrac{1}{3}\) - 1\(\dfrac{3}{4}\)

= \(\dfrac{-7}{3}\) - \(\dfrac{7}{4}\)

= \(\dfrac{-28}{12}\) - \(\dfrac{21}{12}\)

= \(\dfrac{-49}{12}\) 

e; (- \(\dfrac{1}{3}\))(-\(\dfrac{9}{13}\))

=  \(\dfrac{3}{13}\)

f; 1\(\dfrac{1}{2}\) x (- \(\dfrac{10}{9}\))

= \(\dfrac{3}{2}\) x (- \(\dfrac{10}{9}\))

= - \(\dfrac{5}{3}\)

được

\(\dfrac{1}{2}\)\(x^2\) = \(\dfrac{1}{4}x^2\) + \(\dfrac{1}{4}\)\(x^2\)

a: 2x-3=x+1/2

=>\(2x-x=3+\dfrac{1}{2}\)

=>\(x=\dfrac{7}{2}\)

b: \(4x-\left(2x+1\right)=3-\dfrac{1}{3}+x\)

=>\(4x-2x-1=x+\dfrac{8}{3}\)

=>\(2x-1=x+\dfrac{8}{3}\)

=>\(2x-x=\dfrac{8}{3}+1\)

=>\(x=\dfrac{11}{3}\)

\(\dfrac{15}{23}-\dfrac{21}{23}-\left(-\dfrac{8}{23}\right)-\left(-\dfrac{21}{33}\right)\)

\(=\left(\dfrac{15}{23}+\dfrac{8}{23}-\dfrac{21}{23}\right)+\dfrac{21}{33}\)

\(=\dfrac{2}{23}+\dfrac{21}{33}=\dfrac{2\cdot33+21\cdot23}{23\cdot33}=\dfrac{549}{759}\)

Sửa đề: 

\(\dfrac{15}{23}-\dfrac{21}{23}-\left(-\dfrac{8}{23}\right)-\left(-\dfrac{21}{23}\right)\\ =\dfrac{15}{23}-\dfrac{21}{23}+\dfrac{8}{23}+\dfrac{21}{23}\\ =\left(\dfrac{15}{23}+\dfrac{8}{23}\right)+\left(-\dfrac{21}{23}+\dfrac{21}{23}\right)\\ =1+0\\ =1\)

Ta có: \(\left\{{}\begin{matrix}\left|x+1\right|\ge0\\\left(y-2\right)^{2028}\ge0\end{matrix}\right.\) 

=> \(\left|x+1\right|+\left(y-2\right)^{2028}\ge0\)

Dấu = xảy ra khi: 

\(\left\{{}\begin{matrix}x+1=0\\y-2=0\end{matrix}\right.\)

<=> \(\left\{{}\begin{matrix}x=-1\\y=2\end{matrix}\right.\)
Khi đó: 

`A =` \(x^{2024}+\left(5-y\right)^3\)

`=` \(\left(-1\right)^{2024}\) `+ (5-2)^3 `

`= 1 + 3^3 `

`=1 + 27`

`= 28`

Vậy `A = 28`

giúp mình với mình dg vội