Câu trả lời hay nhất:  từ giả thiết thứ nhất dặt x= 3t , y =5t , z = -2t 
thay vào giả thiết thứ 2 ta có 15t - 5t - 6t = 124 <=> t =31 
nên x= 93 , y= 155 , z= -62

thân mên

long

 đặng hoàng long

Ta có (x+1/2)*(2/3-2x)=0

        x+\(\frac{1}{2}\) =0 hoặc \(\frac{2}{3}\) -2x=0

       x     =0-\(\frac{1}{2}\)               2x = \(\frac{2}{3}\)

        x   =\(-\frac{1}{2}\)                x = \(\frac{2}{3}\) : 2 =\(\frac{1}{3}\)

 Vậy x =\(-\frac{1}{2}\)hoặc x = \(\frac{1}{3}\)

TH1 : x=0 =>x=0

TH2: x + 1/2=0 => x= -1/2

TH3: 2/3-2x =0 => x= 1/3

Ta có

(x + \(\frac{1}{2}\))(\(\frac{2}{3}.2x\)) = 0

\(\Rightarrow\)\(\orbr{\begin{cases}\frac{2}{3}.2x=0\\x+\frac{1}{2}=0\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x=0\\x=-\frac{1}{2}\end{cases}}\)

\(a,-2.\left(x+7\right)+3.\left(x-2\right)=-2\)

                \(-2x-14+3x-6=-2\)

                                         \(x-20=-2\)

                                                    \(x=-2+20\)

                                                    \(x=18\)

\(b,-7-2x=-37-\left(-26\right)\)

      \(-7-2x=-11\)

               \(-2x=-11+7\)

               \(-2x=-4\)

                     \(x=2\)

\(c,\left(3x+9\right).\left(11-x\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}3x+9=0\\11-x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-3\\x=11\end{cases}}}\)

\(\orbr{\begin{cases}3x+9=0\\11-x=0\end{cases}}\Rightarrow\orbr{\begin{cases}3x=-9\\x=11\end{cases}}\Rightarrow\orbr{\begin{cases}x=-3\\x=11\end{cases}}\)

Ta có : x = 99

=> 100 = x + 1 

Thay vào A ta có : A = x²⁰¹⁸ - 100x²⁰¹⁷ + 100x²⁰¹⁶ - ...... + 100x² - 100x + 2019

=> A = x²⁰¹⁸ - (x + 1)x²⁰¹⁷ + (x + 1)x²⁰¹⁶ - ...... + (x + 1)x² - (x + 1)x + 2019

=> A = x²⁰¹⁸ - x²⁰¹⁸ - x²⁰¹⁷ + x²⁰¹⁷ + x²⁰¹⁶ -.......+ x³ + x² - x² + x + 2019

=> A = x + 2019

=> A = 99 + 2019

=> A = 2118

P/s : ko cần ! :D 

Theo đề bài ra ta có :

x = 99  

Thay vào A ta có :

A = x²⁰¹⁸ - 100x²⁰¹⁷ + 100x²⁰¹⁶ - ... + 100x² - 100x + 2019

\(\Rightarrow\) A = x²⁰¹⁸ - ( x + 1 ) x²⁰¹⁷ + ( x + 1 ) x²⁰¹⁶ - ... + ( x + 1 ) x² - ( x + 1 ) x + 2019

\(\Rightarrow\) A = x²⁰¹⁸ - x²⁰¹⁸ - x²⁰¹⁷ + x²⁰¹⁷ + x²⁰¹⁶- ... + x³ + x² - x² + x + 2019

\(\Rightarrow\) A = x + 2019

\(\Rightarrow\) A = 99 + 2019

\(\Rightarrow\) A = 2118 

\(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{\left(2x+1\right).\left(2x+3\right)}=\frac{15}{93}\)

\(2.\left(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{\left(2x+1\right).\left(2x+3\right)}\right)=2.\frac{15}{93}\)

\(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{\left(2x+1\right).\left(2x+3\right)}=\frac{30}{93}\)

\(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{2x+1}-\frac{1}{2x+3}=\frac{10}{31}\)

\(\frac{1}{3}-\frac{1}{2x+3}=\frac{10}{31}\)

\(\frac{1}{2x+3}=\frac{1}{3}-\frac{10}{31}\)

\(\frac{1}{2x+3}=\frac{1}{93}\)

=> 2x + 3 = 93

=> 2x = 93 - 3

=> 2x = 90

=> x = 90 : 2

=> x = 45

Vậy x = 45

sai rồi

Đặt \(A=\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{\left(2x+1\right).\left(2x+3\right)}\)

\(2A=\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{\left(2x+1\right).\left(2x+3\right)}\)

\(2A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2x+1}-\frac{1}{2x+3}\)

\(2A=\frac{1}{3}-\frac{1}{2x+3}\)

\(A=\frac{2x}{3\left(2x+3\right)}:2\)

\(\Rightarrow\frac{2x}{3\left(2x+3\right)}:2=\frac{15}{93}\)

\(\frac{2x}{3\left(2x+3\right)}=\frac{15}{93}.2=\frac{30}{93}=\frac{10}{31}\)

\(\frac{2x}{6x+9}=\frac{10}{31}\)

\(\Rightarrow2x.31=10.\left(6x+9\right)\)

\(\Rightarrow62x=60x+90\)

\(62x-60x=90\)

\(2x=90\)

\(x=45\)

Vậy x = 45

Ủng hộ mk nha !!! ^_^