`#3107.101107`

\(\dfrac{1}{4}\times x+\dfrac{4}{5}=-\dfrac{7}{10}\)

\(\Rightarrow\dfrac{1}{4}\times x=-\dfrac{7}{10}-\dfrac{4}{5}\)

\(\Rightarrow\dfrac{1}{4}\times x=-\dfrac{3}{2}\)

\(\Rightarrow x=-\dfrac{3}{2}\div\dfrac{1}{4}\)

\(\Rightarrow x=-6\)

Vậy, `x = -6.`

1/4 . x + 4/5 = -7/10

1/4 . x = -7/10 - 4/5

1/4 . x = -3/2

x = -3/2 : 1/4

x = -6

A = (2/3 - 1/5 + 3a) - 2(1/3 - 1/10 + a)

= 7/15 + 3a - 2.7/30 - 2a

= (3a - 2a) + (7/15 - 7/15)

= a

a) Do BD là tia phân giác của ABC (gt)

⇒ ∠ABD = ∠CBD = 90⁰ : 2 = 45⁰

Do MN // AB (gt)

AB ⊥ BC (gt)

⇒ MN ⊥ BC

⇒ ∠INB = 90⁰

⇒ ∆INB vuông tại N

⇒ ∠BIN = 90⁰ - ∠IBN

= 90⁰ - ∠CBD

= 90⁰ - 45⁰

= 45⁰

⇒ ∠MID = ∠BIN = 45⁰ (đối đỉnh)

b) MN ⊥ BC (đã chứng minh ở câu a)

Tính MID MÀ

A = 2 + 2² + 2³ + ... + 2⁶⁰

= (2 + 2²) + (2³ + 2⁴) + ... + (2⁵⁹ + 2⁶⁰)

= 2.(1 + 2) + 2³.(1 + 2) + ... + 2⁵⁹.(1 + 2)

= 2.3 + 2³.3 + ... + 2⁵⁹.3

= 3.(2 + 2³ + ... + 2⁵⁹) ⋮ 3

Vậy A ⋮ 3

------

A = 2 + 2² + 2³ + ... + 2⁶⁰

= (2 + 2² + 2³) + (2⁴ + 2⁵ + 2⁶) + ... + (2⁵⁸ + 2⁵⁹ + 2⁶⁰)

= 2.(1 + 2 + 2²) + 2⁴.(1 + 2 + 2²) + ... + 2⁵⁸.(1 + 2 + 2²)

= 2.7 + 2⁴.7 + ... + 2⁵⁸.7

= 7.(2 + 2⁴ + ... + 2⁵⁸) ⋮ 7

Vậy A ⋮ 7

--------

A = 2 + 2² + 2³ + ... + 2⁶⁰

= (2 + 2² + 2³ + 2⁴) + (2⁵ + 2⁶ + 2⁷ + 2⁸) + ... + (2⁵⁷ + 2⁵⁸ + 2⁵⁹ + 2⁶⁰)

= 30 + 2⁴.(2 + 2² + 2³ + 2⁴) + ... + 2⁵⁶.(2 + 2² + 2³ + 2⁴)

= 30.(1 + 2⁴ + ... + 2⁵⁶)

= 5.6.(1 + 2⁴ + ... + 2⁵⁶) ⋮ 5

Vậy A ⋮ 5

\(A=2+2^2+2^3+...+2^{60}\)

\(A=\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{59}+2^{60}\right)\)

\(A=6+2^2.\left(2+2^2\right)+...+2^{58}.\left(2+2^2\right)\)

\(A=6+2^2.6+...+2^{58}.6\)

\(A=6.\left(1+2^2+...+2^{58}\right)\) 

Vì \(6⋮3\) nên \(6.\left(1+2^2+...+2^{58}\right)⋮3\)

Vậy \(A⋮3\)

___________

\(A=2+2^2+2^3+...+2^{60}\)

\(A=\left(2+2^2+2^3\right)+...+\left(2^{58}+2^{59}+2^{60}\right)\)

\(A=14+...+2^{57}.\left(2+2^2+2^3\right)\)

\(A=14+...+2^{57}.14\)

\(A=14.\left(1+...+2^{57}\right)\)

Vì \(14⋮7\) nên \(14.\left(1+...2^{57}\right)⋮7\)

Vậy \(A⋮7\)

____________

\(A=2+2^2+2^3+...+2^{60}\)

\(A=\left(2+2^2+2^3+2^4\right)+...+\left(2^{57}+2^{58}+2^{59}+2^{60}\right)\)

\(A=30+...+2^{56}.\left(2+2^2+2^3+2^4\right)\)

\(A=30+...+2^{56}.30\)

\(A=30.\left(1+...+2^{56}\right)\)

Vì \(30⋮5\) nên \(30.\left(1+...+2^{56}\right)⋮5\)

Vậy \(A⋮7\)

\(#WendyDang\)

Bạn cần ghi đầy đủ yêu cầu đề và điều kiện của $x,y$ để được hỗ trợ tốt hơn nhé.

2xy-2x+3y=21

=>2x(y-1)+3y-3=18

=>(y-1)(2x+3)=18

=>\(\left(2x+3;y-1\right)\in\left\{\left(1;18\right);\left(2;9\right);\left(3;6\right);\left(6;3\right);\left(9;2\right);\left(18;1\right);\left(-1;-18\right);\left(-2;-9\right);\left(-3;-6\right);\left(-6;-3\right);\left(-9;-2\right);\left(-18;-1\right)\right\}\)

=>\(\left(x,y\right)\in\left\{\left(-1;19\right);\left(-\dfrac{1}{2};10\right);\left(0;7\right);\left(\dfrac{3}{2};4\right);\left(3;3\right);\left(\dfrac{15}{2};2\right);\left(-2;-17\right);\left(-\dfrac{5}{2};-4\right);\left(-3;-5\right);\left(-\dfrac{9}{2};-2\right);\left(-6;-1\right);\left(-\dfrac{21}{2};0\right)\right\}\)

Ta có:

\(\left(-16\right)^{11}=\left(-2^4\right)^{11}=\left(-2\right)^{4.11}=\left(-2\right)^{44}\)

\(\left(-32\right)^9=\left(-2^5\right)^9=\left(-2\right)^{5.9}=\left(-2\right)^{45}\)

Vì \(44< 45\) nên \(\left(-2\right)^{44}>\left(-2\right)^{45}\)

Vậy \(\left(-16\right)^{11}>\left(-32\right)^9\)

\((9-x)^2-7\)

Ta thấy: \(\left(9-x\right)^2\ge0\forall x\)

\(\Rightarrow\left(9-x\right)^2-7\ge-7\forall x\)

Dấu \("="\) xảy ra khi: \(9-x=0\Leftrightarrow x=9\)

Vậy GTNN của biểu thức là -7 khi x = 9.

`2x -2/3 +1/2x =-1`

`=> 2x+1/2x =-1+2/3`

`=> (2+1/2) x =-3/3 + 2/3`

`= 5/2 x = -1/3`

`=> x=-1/3 : 5/2`

`=>x= -2/15`

`31/36 - (1/3-x)^2 =5/6`

`=>  (1/3-x)^2 = 31/36 - 5/6`

`=>  (1/3-x)^2 =1/36`

`=>  (1/3-x)^2 = (+-1/6)^2`

\(\Rightarrow\left[{}\begin{matrix}\dfrac{1}{3}-x=\dfrac{1}{6}\\\dfrac{1}{3}-x=-\dfrac{1}{6}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{6}\\x=\dfrac{1}{2}\end{matrix}\right.\)