\(\dfrac{x+5}{2017}+\dfrac{x+4}{2018}+\dfrac{x+3}{2019}=-3\\ \dfrac{x+5}{2017}+1+\dfrac{x+4}{2018}+1+\dfrac{x+3}{2019}=-3+3\\ \dfrac{x+5}{2017}+\dfrac{2017}{2017}+\dfrac{x+4}{2018}+\dfrac{2018}{2018}+\dfrac{x+3}{2019}+\dfrac{2019}{2019}=0\\ \dfrac{x+2022}{2017}+\dfrac{x+2022}{2018}+\dfrac{x+2022}{2019}=0\\ x+2022.\left(\dfrac{1}{2017}+\dfrac{1}{2018}+\dfrac{1}{2019}\right)=0\)

⇒x+2022=0 (vì \(\dfrac{1}{2017}+\dfrac{1}{2018}+\dfrac{1}{2019}\)\(\ne0\))

⇒x=0-2022

⇒x=-2022

a) Đáp án là B

b) Đáp án là C

Câu A đáp án đúng là B

Câu B đáp án đúng là C

vậy nha bạn chúc bạn học tốt 

CẢM ƠN TẤT CẢ CÁC BẠN NHÉ !

a) 

\(\frac{7}{4}:\left(\frac{2}{3}-\frac{5}{4}\right).\left(\frac{-1}{4}\right)=\frac{7}{4}:\frac{-7}{12}.\left(\frac{-1}{4}\right)=-3.\left(\frac{-1}{4}\right)=\frac{3}{4}\)

b) \(\frac{2}{-3x}=\frac{4}{5}\)

\(\Rightarrow\frac{4}{-6x}=\frac{4}{5}\)

\(\Rightarrow-6x=5\)

\(\Rightarrow x=\frac{-5}{6}\)

Vậy \(x=\frac{-5}{6}\)

\(\left\{x^2-\left[6^2-\left(8^2-9.7\right)^3-7.5\right]^3-5.3\right\}^3=1\)

\(\Rightarrow\left\{x^2-\left(36-1^3-35\right)^3-15\right\}^3=1\)

\(\Rightarrow x^2-\left(0^3-15\right)^3=1\)

\(\Rightarrow x^2-\left(-3375\right)=1\)

\(\Rightarrow x^2=-3374\)

\(\Rightarrow x\in\varnothing\)

{ x² - [ 6² - ( 8² - 9.7)³ - 7.5]³ - 5.3 }³ = 1

{ x² + [ 36 - (64 - 63)³ - 35]³ - 15}³ = 1

[ x² - ( 36 - 1³ - 35 ) - 15 ]³ = 1

[ x² - ( 36 - 1 - 35 ) - 15]³ = 1

[ x² - ( 35 - 35 ) - 15]³ = 1

[ x² - 0 - 15]³ = 1

( x² - 15 )³ = 1

<=> ( x² - 15)³ = 1³

=> x² - 15 = 1

<=> x² = 16

=> x = 4

Ta có \(\left(x-3\right)^4=\left(x-3\right)^6\)

\(\Rightarrow\left(x-3\right)^6-\left(x-3\right)^4=0\)

\(\Rightarrow\left(x-3\right)^4.\left[\left(x-3\right)^2-1\right]=0\)

\(\Rightarrow\left(x-3\right)^4=0\)hoặc \(\left(x-3\right)^2-1=0\)

Với \(\left(x-3\right)^4=0\Rightarrow x-3=0\Rightarrow x=3\)

Với \(\left(x-3\right)^2-1=0\Rightarrow\left(x-3\right)=1\Rightarrow\orbr{\begin{cases}x-3=1\\x-3=-1\end{cases}}\Rightarrow\orbr{\begin{cases}x=4\\x=2\end{cases}}\)

Vậy \(x\in\left\{2;3;4\right\}\)

<=>(x-3)^4-(x-3)^6=0

<=>(x-3)4×[1-(x-3)^2]=0

<=>(x-3)^4=0 hoặc 1-(x-3)^2=0

Sau đó bn lập bảng mà tìm nha !

Tk mình đi!

Ai Lm đúng và Trả lời nhanh nhất tớ tik !!

A=(-3/4+2/3).11/9+(-1/4+1/3):|-9/11|

\(=\left(-\frac{3}{4}+\frac{2}{3}\right).\frac{11}{9}+\left(-\frac{1}{4}+\frac{1}{3}\right).\frac{11}{9}\)

\(=\frac{11}{9}\left(-\frac{3}{4}+\frac{2}{3}-\frac{1}{4}+\frac{1}{3}\right)\)

\(=\frac{11}{9}\left(-1+1\right)\)

\(=\frac{11}{9}.0\)

=0

\(M\left(x\right)=-3x^2+6x-4+2x^2-5x+4=-x^2+x\)

Đặt M(x)=0

=>-x(x-1)=0

=>x=0 hoặc x=1

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

Giả sử: \(M\left(x\right)=0\)

\(\Leftrightarrow-x\left(x-1\right)=0\)

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