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\(1.\left(x-5\right)^{23}.\left(y+2\right)^7=0\)

\(\Rightarrow\hept{\begin{cases}\left(x-5\right)^{23}=0\\\left(y+2\right)^7=0\end{cases}\Rightarrow\hept{\begin{cases}\left(x-5\right)^{23}=0^{23}\\\left(y+2\right)^7=0^7\end{cases}}}\)\(\Rightarrow\hept{\begin{cases}x-5=0\\y+2=0\end{cases}\Rightarrow\hept{\begin{cases}x=0+5\\y=0-2\end{cases}}}\)\(\Rightarrow\hept{\begin{cases}x=5\\y=-2\end{cases}}\)

Vậy \(\left(x;y\right)=\left(5;-2\right)\)

2. \(A=\left(x-2\right)^2+|y+3|+7\)

Ta có :

\(\hept{\begin{cases}\left(x-2\right)^2\ge0\forall x\\|y+3|\ge0\forall y\end{cases}}\)

\(\Rightarrow\left(x-2\right)^2+|y+3|\ge0\forall x;y\)

\(\Rightarrow\left(x-2\right)^2+|y+3|+7\ge7\forall x;y\)

\(\Rightarrow A\ge7\forall x;y\)

Dấu bằng xảy ra

\(\Leftrightarrow\hept{\begin{cases}\left(x-2\right)^2=0\\|y+3|=0\end{cases}\Leftrightarrow\hept{\begin{cases}x-2=0\\y+3=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x=2\\y=-3\end{cases}}}\)

Vậy GTNN của A là 7 khi \(\left(x;y\right)=\left(2;-3\right)\)

Bài 1:

a) \(=\dfrac{8}{15}\left(\dfrac{7}{13}+\dfrac{6}{13}\right)=\dfrac{8}{15}.1=\dfrac{8}{15}\)

b) \(=\dfrac{3.3-7-2.4}{12}=-\dfrac{6}{12}=-\dfrac{1}{2}\)

Bài 2:

 \(\dfrac{x}{2,7}=-\dfrac{2}{3,6}\Rightarrow x=\dfrac{\left(-2\right).2,7}{3,6}\Rightarrow x=-\dfrac{3}{2}\)

Bài 3:

\(\dfrac{x}{2}=\dfrac{y}{5}=\dfrac{x+y}{2+5}=-\dfrac{21}{7}=-3\)

\(\Rightarrow\left\{{}\begin{matrix}x=\left(-3\right).2=-6\\y=\left(-3\right).5=-10\end{matrix}\right.\)

=>7^x*49+7^x*2=51

=>7^x*51=51

=>7^x=1

=>x=0

\(7^{x+2}+7^x\cdot2=51\\ \Leftrightarrow7^x\cdot7^2+7^x\cdot2=51\\ \Leftrightarrow7^x\cdot51=51\\ \Leftrightarrow7^x=1\\ \Leftrightarrow x=0\)

Lời giải:

$49x=|2x+7|+|2x+7^2|+....+|2x+7^{50}|\geq 0$

$\Rightarrow x\geq 0$

$\Rightarrow 2x+7>0; 2x+7^2>0;....; 2x+7^{50}>0$

Do đó bài toán trở thành:

$(2x+7)+(2x+7^2)+....+(2x+7^{50})=49x$

$\underbrace{(2x+2x+...+2x)}_{50}+(7+7^2+....+7^{50})=49x$

$\Rightarrow 100x+(7+7^2+....+7^{50})=49x$

$\Rightarrow 7+7^2+....+7^{50} = -51x>0$

$\Rightarrow x<0$ (vô lý - loại) 

Vậy không tồn tại $x$ thỏa mãn đề.

5 ( x - 1 ) - 7 ( x - 2 ) = 2x - 39

<=> 5x - 5 - 7x + 14 = 2x - 39

<=> 5x - 7x - 2x = -39 + 5 - 14

<=> -4x = -48

<=> x = 12

x - 3 - 14.( x-2 )= -3x -3\(\Rightarrow\chi-3-28-14\chi-28=-3\chi-3\)

\(\Rightarrow\chi-3-28+3=-3\chi-3\)

\(\Rightarrow\chi-28=11\chi\)

\(\Rightarrow\chi-11\chi=28\)

\(\Rightarrow10\chi=28\Rightarrow\chi=2,8\left(kot.m\chi\inℤ\right)\) 

HTDT-sai-rùi

x-3-14(x-2)=-3x-3

x-3+28-14x=-3x-3

-13x+35=-3x-3

-13x+3x=-3-35

-10x=-38

x=38/10=19/5

Bài 1:

a) \(\dfrac{9}{20}-\dfrac{8}{15}\times\dfrac{5}{12}\)

\(=\dfrac{9}{20}-\dfrac{2}{9}\)

\(=\dfrac{41}{180}\)

b) \(\dfrac{2}{3}\div\dfrac{4}{5}\div\dfrac{7}{12}\)

\(=\dfrac{2}{3}\times\dfrac{5}{4}\times\dfrac{12}{7}\)

\(=\dfrac{5}{6}\times\dfrac{12}{7}\)

\(=\dfrac{10}{7}\)

c) \(\dfrac{7}{9}\times\dfrac{1}{3}+\dfrac{7}{9}\times\dfrac{2}{3}\)

\(=\dfrac{7}{9}\times\left(\dfrac{1}{3}+\dfrac{2}{3}\right)\)

\(=\dfrac{7}{9}\times1\)

\(=\dfrac{7}{9}\)

Bài 2:

a) \(2\times\left(x-1\right)=4026\)

\(\left(x-1\right)=4026\div2\)

\(x-1=2013\)

\(x=2014\)

Vậy: \(x=2014\)

b) \(x\times3,7+6,3\times x=320\)

\(x\times\left(3,7+6,3\right)=320\)

\(x\times10=320\)

\(x=320\div10\)

\(x=32\)

Vậy: \(x=32\)

c) \(0,25\times3< 3< 1,02\)

\(\Leftrightarrow0,75< 3< 1,02\) ( S )

=> \(0,75< 1,02< 3\)

( 3x - 2⁴ ) . 7³ = 2 . 7⁴

3x - 16 = 2 . 7⁴ : 7³ = 14

3x = 14 + 16 = 30

x = 10

\(\left(3\times x-2^4\right)\times7^3=2\times7^4\)

 \(\left(3\times x-2^4\right)\div2=7^4\div7^3\) 

  \(\left(3\times x-16\right)\div2=7\)

           \(3\times x-16=7\times2\) 

           \(3\times x-16=14\) 

                     \(3\times x=14+16\)                      

                      \(3\times x=30\)

                           \(x=30\div3\) 

                           \(x=10\)

x = 10