Em có nhầm đề ko nhỉ lớp 7 ko giải được bài này

\(Sai\) \(Thầy\) \(Ơi\)

a; \(x^3\) + 64 

= \(x^3\) + 4³

= (\(x+4\))(\(x^2\) - 4\(x\) + 16)

b; 2\(x^2\) - 4\(x\)

= 2\(x\)(\(x-2\))

c; 6\(x^2\)y + 4\(xy^2\) + 2\(xy\)

= 2\(xy\)(3\(x\) + 2y + 1)

a) x³ + 64

= x³ + 4³

= (x + 4)(x² − 4x + 16)

b) 2x² − 4x

= 2x(x - 2)

c) 6x²y + 4xy² + 2xy

= 2xy(3x + 2y + 1)

d) Sửa đề: x² − x + y − 2xy + y²

= x² − 2xy + y² − x + y

= (x − y)² − (x − y)

= (x − y)(x − y − 1)

3,2>3,1>2,8

vì 2,8 < 3,11 < 3,2 Vậy số lớn nhất là số 3,2

\(P=\left(-\dfrac{1}{7}\right)^0+\left(-\dfrac{1}{7}\right)^1+\left(-\dfrac{1}{7}\right)^2+...+\left(-\dfrac{1}{7}\right)^{2017}\)

\(\left(-\dfrac{1}{7}\right).P=\left(-\dfrac{1}{7}\right)^1+\left(-\dfrac{1}{7}\right)^2+...+\left(-\dfrac{1}{7}\right)^{2017}+\left(-\dfrac{1}{7}\right)^{2018}\)

\(P-\left(-\dfrac{1}{7}\right)P=\left(-\dfrac{1}{7}\right)^0-\left(-\dfrac{1}{7}\right)^{2018}\)

\(\dfrac{8}{7}P=1-\dfrac{1}{7^{2018}}\)

\(\dfrac{8}{7}P=\dfrac{7^{2018}-1}{7^{2018}}\)

\(P=\dfrac{7^{2018}-1}{8.7^{2017}}\)

A = (\(x+y\))² - 2.(\(x+y\))z + 4z²

A = (\(x+y\))² - 2.(\(x+y\))z + (2z)²

A = (\(x+y\) - 2z)²

A = (\(x+y\) - 2z)(\(x+y\) - 2z)

\(B=\left(\dfrac{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}+a\right)}{1-\sqrt{a}}+\sqrt{a}\right)\left(\dfrac{1-\sqrt{a}}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)}\right)^2\)

\(=\left(a+2\sqrt{a}+1\right)\left(\dfrac{1}{1+\sqrt{a}}\right)^2\)

\(=\left(\sqrt{a}+1\right)^2.\dfrac{1}{\left(\sqrt{a}+1\right)^2}=1\)

\(C=\dfrac{\sqrt{x}-\sqrt{y}}{xy\sqrt{xy}}:\left[\left(\dfrac{x+y}{xy}\right).\dfrac{1}{\left(\sqrt{x}+\sqrt{y}\right)^2}+\dfrac{2}{\left(\sqrt{x}+\sqrt{y}\right)^3}.\left(\dfrac{\sqrt{x}+\sqrt{y}}{\sqrt{xy}}\right)\right]\)

\(=\dfrac{\sqrt{x}-\sqrt{y}}{xy\sqrt{xy}}:\left[\dfrac{x+y}{xy\left(\sqrt{x}+\sqrt{y}\right)^2}+\dfrac{2}{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)^2}\right]\)

\(=\dfrac{\sqrt{x}-\sqrt{y}}{xy\sqrt{xy}}:\left[\dfrac{x+y}{xy\left(\sqrt{x}+\sqrt{y}\right)^2}+\dfrac{2\sqrt{xy}}{xy\left(\sqrt{x}+\sqrt{y}\right)^2}\right]\)

\(=\dfrac{x-y}{xy\sqrt{xy}}:\left[\dfrac{x+y+2\sqrt{xy}}{xy\left(\sqrt{x}+\sqrt{y}\right)^2}\right]\)

\(=\dfrac{x-y}{xy\sqrt{xy}}:\left[\dfrac{\left(\sqrt{x}+\sqrt{y}\right)^2}{xy\left(\sqrt{x}+\sqrt{y}\right)^2}\right]=\dfrac{x-y}{xy\sqrt{xy}}.xy\)

\(=\dfrac{x-y}{\sqrt{xy}}\)

\(=\dfrac{2-\sqrt{3}-\left(2+\sqrt{3}\right)}{\sqrt{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}}=\dfrac{-2\sqrt{3}}{\sqrt{4-3}}=-2\sqrt{3}\)

\(A=\dfrac{\sqrt{2}\left(1+\sqrt{5}\right)}{2+\sqrt{6+2\sqrt{5}}}+\dfrac{\sqrt{2}\left(1-\sqrt{5}\right)}{2-\sqrt{6-2\sqrt{5}}}\)

\(=\dfrac{\sqrt{2}\left(1+\sqrt{5}\right)}{2+\sqrt{\left(\sqrt{5}+1\right)^2}}+\dfrac{\sqrt{2}\left(1-\sqrt{5}\right)}{2-\sqrt{\left(\sqrt{5}-1\right)^2}}\)

\(=\dfrac{\sqrt{2}\left(1+\sqrt{5}\right)}{2+\sqrt{5}+1}+\dfrac{\sqrt{2}\left(1-\sqrt{5}\right)}{2-\left(\sqrt{5}-1\right)}\)

\(=\dfrac{\sqrt{2}\left(1+\sqrt{5}\right)}{3+\sqrt{5}}+\dfrac{\sqrt{2}\left(1-\sqrt{5}\right)}{3-\sqrt{5}}\)

\(=\sqrt{2}\left(\dfrac{\left(1+\sqrt{5}\right)\left(3-\sqrt{5}\right)+\left(1-\sqrt{5}\right)\left(3+\sqrt{5}\right)}{\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)}\right)\)

\(=\sqrt{2}.\left(\dfrac{-4}{9-5}\right)=-\sqrt{2}\)

                Giải:

Sau khi có khác lên tàu và xuồn tàu thì số khách trên tàu giảm đi là:

       158 - 127 = 31 (khách)

Ban đầu tàu có số khách là:

        191 + 31  = 222 (khách)

Đáp số: 222 khách 

               Giải:

Đổi 2 tấn 4 tạ: = 2400 kg

Tổng số xe tham gia chở cám là:

               3 + 4 = 7 (xe)

Tổng số cám mà cả đoàn xe đã chở là:

    2400 x 3 + 2820 x 4 =  18480 (kg)

Trung bình mỗi xe chở được số ki-lô-gam cám là:

       18480 : 7 = 2640 (kg)

Đáp số:... 

\(\dfrac{x-1}{69}+\dfrac{x-2}{68}=\dfrac{x-5}{65}+\dfrac{x-6}{64}\)

\(\Leftrightarrow\left(\dfrac{x-1}{69}-1\right)+\left(\dfrac{x-2}{68}-1\right)-\left(\dfrac{x-5}{65}-1\right)-\left(\dfrac{x-6}{64}-1\right)=0\)

\(\Leftrightarrow\dfrac{x-70}{69}+\dfrac{x-70}{68}-\dfrac{x-70}{65}-\dfrac{x-70}{64}=0\)

\(\Leftrightarrow\left(x-70\right)\left(\dfrac{1}{69}+\dfrac{1}{68}-\dfrac{1}{65}-\dfrac{1}{64}\right)=0\)

\(\Leftrightarrow x-70=0\) (do \(\dfrac{1}{69}+\dfrac{1}{68}-\dfrac{1}{65}-\dfrac{1}{64}\ne0\))

\(\Rightarrow x=70\)

      A = 3 + 3² + 3³ + ... + 3²⁰¹⁵

     3A = 3² + 3³ + 3⁴ + ... + 3²⁰¹⁶

3A - A = 3² + 3³ + 3⁴ + ... + 3²⁰¹⁶

    2A    = 3² + 3³ + 3⁴ + ... + 3²⁰¹⁶ - (3 + 3² + 3³ + ... + 3²⁰¹⁵)

    2A =  3² + 3³ + 3⁴ + ... + 3²⁰¹⁶ - 3 - 3² - 3³ - ... - 3²⁰¹⁵

  2 A =  3 ²⁰¹⁶ - 3 + (3² - 3²) + (3³ - 3³) + ... + (3²⁰¹⁵ - 3²⁰¹⁵)

    2A =  3²⁰¹⁶ - 3 + 0 + 0 + ... + 0 + 0

    2A = 3²⁰¹⁶ - 3

2A + 3 = 3²⁰¹⁶ - 3 + 3 

 2A + 3 = 3²⁰¹⁶ - (3 - 3) = 3^4\(x\)

3²⁰¹⁶ = 3^4\(x\)

2016 = 4\(x\)

  \(x\) = 2016 : 4

  \(x=\) 504

Vậy \(x=504\)