HOC24 có câu rất hay :Người hay giúp bạn khác trả lời bài tập sẽ trở thành học sinh giỏi. Người hay hỏi bài thì không. Còn bạn thì sao? đúng tính bà đó . Lên lớp đừng đập nha :)

a) 3 . ( 1/2 - x ) + 1/3 = 7/6 - x

=> 3/2 - 3x + 1/3 = 7/6-x

=> -3x +x=7/6 - 3/2 - 1/3

=> -2x = -2/3

=> x=-2/3 : (-2) = 1/3

hết :)

\(x\times\frac{3}{5}+x:5=\frac{2}{3}\)

\(x\times\frac{3}{5}+x\times\frac{1}{5}=\frac{2}{3}\)

\(\frac{3}{5}x+\frac{1}{5}x=\frac{2}{3}\)

\(\frac{4}{5}x=\frac{2}{3}\)

\(x=\frac{4}{5}:\frac{2}{3}\)

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

Phần b tương tự.

mot oto di mot quang duong trong 3 gio gio thu nhat di duoc 1/4 quang duong , gio thu hai di duoc 2/5 quang duong do , Hoi gio thu ba oto di duoc may phan quang duong do ?

1: 

\(\Leftrightarrow\left(x^2+5x+6\right)\left(x^2+5x+4\right)=24\)

\(\Leftrightarrow\left(x^2+5x\right)^2+10\left(x^2+5x\right)=0\)

\(\Leftrightarrow x^2+5x=0\)

=>x=0 hoặc x=-5

3: \(\Leftrightarrow\left(x^2+x+6\right)\left(x^2+x-2\right)=0\)

=>(x+2)(x-1)=0

=>x=-2 hoặc x=1

\(A=\left(x+1\right)^3-\left(x+3\right)^2\left(x+1\right)+4x^2+8\)

\(A=x^3+3x^2+3x+1-\left(x^2+6x+9\right)\left(x+1\right)+4x^2+8\)

\(A=x^3+3x^2+3x+1-\left(x^3+6x^2+9x+x^2+6x+9\right)+4x^2+8\)

\(A=x^3+3x^2+3x+1-x^3-6x^2-9x-x^2-6x-9+4x^2+8\)

\(A=\left(x^3-x^3\right)+\left(3x^2-6x^2-x^2+4x^2\right)+\left(3x-9x-6x\right)+\left(1-9+8\right)\)

\(A=-12x\)

\(B=\left(x-2\right)\left(x^2+2x+4\right)-\left(x+1\right)^3+3\left(x-1\right)\left(x+1\right)\)

\(B=x^3+2x^2+4x-2x^2-4x-8-\left(x^3+3x^2+3x+1\right)+3\left(x^2-1\right)\)

\(B=x^3+2x^2+4x-2x^2-4x-8-x^3-3x^2-3x-1+3x^2-3\)

\(B=\left(x^3-x^3\right)+\left(2x^2-2x^2-3x^2+3x^2\right)+\left(4x-4x-3x\right)+\left(-8-3-1\right)\)

\(B=-3x-12\)

Câu C tương tự.

Chúc bạn học tốt!!!

A = \(\left(x+1\right)^3-\left(x+3\right)^2.\left(x+1\right)+4x^2+8\)

A = \(\left(x+1\right)\left(x+1-x-3\right)\left(x+1+x+3\right)+4x^2+8\)

A = \(\left(x+1\right).\left(-2\right).\left(2x+4\right)+4x^2+8\)

A = \(\left(-2\right)\left(2x^2+4x+2x+4\right)+4x^2+8\)

A = \(\left(-2\right)\left(2x^2+6x+4\right)+4x^2+8\)

A = \(-4x^2-12x-8+4x^2+8=-12x\)

b) B = \(\left(x-2\right)\left(x^2+2x+4\right)-\left(x+1\right)^3+3\left(x-1\right)\left(x+1\right)\)

B = \(x^3-8-\left(x+1\right)\left(x^2+2x+1+3x-3\right)\)

B = \(x^3-8-\left(x+1\right)\left(x^2+5x-2\right)\)

B = \(x^3-8-x^3-5x^2+2x-x^2-5x+2\)

B = \(-6x^2-3x-6\)

3/5 X + 2X + 40 = 53

 13/5 X              = 53 - 40

 13/5 X              = 13

    X                   = 13 : 13/5

    X                  = 5

\(\frac{3}{5}x+2x+40=53\)

\(\Leftrightarrow\frac{13}{5}x+40=53\)

\(\Leftrightarrow\frac{13}{5}x=53-40\)

\(\Leftrightarrow\frac{13}{5}x=13\)

\(\text{Mẫu chung:5}\)

\(\Leftrightarrow13.5x=13.5\)

\(65x=65\)

\(\Rightarrow x=1\)

Bạn @Diedevil sai rồi @@ Thử lại là biết

Ra tới cuối là phép nhân cần gì mẫu chung trời??

( x + 1) + )+ ( x + 2) +( x + 3) +) + ( x + 4 ) + ( x + 5 ) = 40

( x+x+x+x+x ) + ( 1+2+3+4+5 ) = 40

x*5 + 15 = 40

x*5 = 40-15 

x*5 = 25

x= 25 :5

x=5

x+1+x+2+x+3+x+4+x+5=40

5*x+(1+2+3+4+5)         =40

5*x+15                          =40

5*x                                =40-15

5*x                                =25

x                                   =25/5

x                                   =5

1)

\(\left(x+1\right)\left(x+2\right)\left(x+4\right)\left(x+5\right)=40\)

\(\Leftrightarrow\left(x+1\right)\left(x+5\right).\left(x+2\right)\left(x+4\right)-40=0\)

\(\Leftrightarrow\left(x^2+6x+5\right).\left(x^2+6x+8\right)-40=0\)

Đặt \(a=x^2+6x+6\) ta có:

\(\Leftrightarrow\left(a-1\right)\left(a+2\right)-40=0\)

\(\Leftrightarrow a^2+a-2-40=0\)

\(\Leftrightarrow a^2-6x+7x-42=0\)

\(\Leftrightarrow a\left(a-6\right)+7\left(a-6\right)=0\)

\(\Leftrightarrow\left(a-6\right)\left(a+7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}a=6\\a=-7\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2+6x+6=6\\x^2+6x+6=-7\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2+6x=0\\x^2+6x+13=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-6\\x=0\end{matrix}\right.\)

(\(x^2+6x+13=\left(x+3\right)^2+4>0\left(loại\right)\))

Vậy.................

3)

\(\left|x+4\right|=\left|3-2x\right|\)

\(\Leftrightarrow\left[{}\begin{matrix}x+4=3-2x\\x+4=-3+2x\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}3x+1=0\\-x+7=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-1}{3}\\x=7\end{matrix}\right.\)

Vậy..........

2)\(x^3-7x^2+15x-25=0\)

\(\Leftrightarrow x^3-5x^2-2x^2+10x+5x-25=0\)

\(\Leftrightarrow x^2\left(x-5\right)-2x\left(x-5\right)+5\left(x-5\right)=0\)

\(\Leftrightarrow\left(x-5\right)\left(x^2-2x+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\x^2-2x+5=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=5\\\left(x-1\right)^2+4=0\left(loai\right)\end{matrix}\right.\)