1C

2A

1C        2A

Bài 1: 

a) Ta có: \(\dfrac{17}{6}-x\left(x-\dfrac{7}{6}\right)=\dfrac{7}{4}\)

\(\Leftrightarrow\dfrac{17}{6}-x^2+\dfrac{7}{6}x-\dfrac{7}{4}=0\)

\(\Leftrightarrow-x^2+\dfrac{7}{6}x+\dfrac{13}{12}=0\)

\(\Leftrightarrow-12x^2+14x+13=0\)

\(\Delta=14^2-4\cdot\left(-12\right)\cdot13=196+624=820\)

Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:

\(\left\{{}\begin{matrix}x_1=\dfrac{14-2\sqrt{205}}{-24}=\dfrac{-7+\sqrt{205}}{12}\\x_2=\dfrac{14+2\sqrt{2015}}{-24}=\dfrac{-7-\sqrt{205}}{12}\end{matrix}\right.\)

b) Ta có: \(\dfrac{3}{35}-\left(\dfrac{3}{5}-x\right)=\dfrac{2}{7}\)

\(\Leftrightarrow\dfrac{3}{5}-x=\dfrac{3}{35}-\dfrac{10}{35}=\dfrac{-7}{35}=\dfrac{-1}{5}\)

hay \(x=\dfrac{3}{5}-\dfrac{-1}{5}=\dfrac{3}{5}+\dfrac{1}{5}=\dfrac{4}{5}\)

ai giúp mik vs

1.

$x(x+2)(x+4)(x+6)+8$

$=x(x+6)(x+2)(x+4)+8=(x^2+6x)(x^2+6x+8)+8$

$=a(a+8)+8$ (đặt $x^2+6x=a$)

$=a^2+8a+8=(a+4)^2-8=(x^2+6x+4)^2-8\geq -8$

Vậy $A_{\min}=-8$ khi $x^2+6x+4=0\Leftrightarrow x=-3\pm \sqrt{5}$

2.

$B=5+(1-x)(x+2)(x+3)(x+6)=5-(x-1)(x+6)(x+2)(x+3)$

$=5-(x^2+5x-6)(x^2+5x+6)$

$=5-[(x^2+5x)^2-6^2]$

$=41-(x^2+5x)^2\leq 41$

Vậy $B_{\max}=41$. Giá trị này đạt tại $x^2+5x=0\Leftrightarrow x=0$ hoặc $x=-5$

bài 1 : a,ta có 3/x-1 =4/y-2=5/z-3 =>  x-1/3=y-2/4=z-3/5 

áp dụng .... => x-1+y-2+z-3 / 3+4+5 = x+y+z-1-2-3/3+4+5 = 12/12=1

do x-1/3 = 1 => x-1 = 3 => x= 4 ( tìm y,z tương tự

Bài 1: 

a) Ta có: 3/x - 1 = 4/y - 2 = 5/z - 3 => x - 1/3 = y - 2/4 = z - 3/5 áp dụng ... =>x - 1 + y - 2 + z - 3/3 + 4 + 5 = x + y + z - 1 - 2 - 3/3 + 4 + 5 = 12/12 = 1 do x - 1/3 = 1 => x - 1 = 3 => x = 4 ( tìm y, z tương tự )

MẤY DÒNG NÀO BẠN THẤY KO CẦN THIẾT THÌ LƯỢC BỎ NHA!!!

a) \(2\left(x-5\right)-3\left(x+6\right)=4\left(x-7\right)\)

   \(2x-10-3x-18=4x-28\)

   \(2x-3x-4x-10-18=-28\)

   \(-5x-28=-28\)

   \(-5x=-28+28=0\)

    \(x=\frac{0}{-5}=0\)

b) \(3\left(x-1\right)-2\left(x+5\right)=2\left(x-3\right)\)

  \(3x-3-2x-10=2x-6\)

   \(3x-2x-2x-3-10=-6\)

   \(-x-13=-6\)

   \(-x=-6+13=7\)

    \(x=-7\)

c) ​​\(5\left(1-x\right)-6\left(1+x\right)=7\left(3-x\right)\)

    ​​\(5-5x-6-6x=21-7x\)

    \(-5x-6x+7x+5-6=21\)

    \(-4x-1=21\)

    \(-4x=22\)

     \(x=\frac{22}{-4}=\frac{-11}{2}\)

d) \(2x+5-3\left(3x+7\right)=6\left(1-x\right)+8\)

    \(2x+5-9x-21=6-6x+8\)

    \(2x-9x+6x+5-21=6+8\)

    ​\(-x-16=14\)

    \(-x=14+16=30\)

    \(x=-30\)

e) \(x-2+3\left(x-4\right)=5\left(x-6\right)+7\)

   \(x-2+3x-12=5x-30+7\)

   \(x+3x-5x-2-12=-30+7\)

  \(-x-14=-23\)

  \(-x=-23+14=-9\)

​  \(x=9\)

f) \(x+2+3\left(1-x\right)-5\left(2-x\right)=6\left(1-x\right)+\left(3-x\right)\)

  \(x+2+3-3x-10+5x=6-6x+3-x\)

  \(x-3x+5x+6x+x+2+3-10=6+3\)

  \(10x-7=9\)

  \(10x=9+7=16\)

  \(x=\frac{16}{10}=\frac{8}{5}\)

 720 : ( x . 2 + x . 3 ) = 3.2
720 : ( x . 2 + x.3 ) = 6
( x .2 + x.3 )           = 720 : 6 
x.2+x.3 = 120
x . ( 2 + 3 ) = 120
x . 5 = 120
     x     = 120 : 5 
    x      = 24

a) \(\left(x+1\right)^3-\left(x-1\right)^3-6\cdot\left(x-1\right)^2=10\)

\(\Rightarrow x^3+3x^2+3x+1-x^3+3x^2-3x+1-6\cdot\left(x^2-2x+1\right)=10\)

\(\Rightarrow6x^2+2-6x^2+12x-6=10\)

\(\Rightarrow12x-4=10\)

\(\Rightarrow12x=14\)

\(\Rightarrow x=\dfrac{7}{6}\)

b) \(x\left(x+5\right)\left(x-5\right)-\left(x+2\right)\left(x^2-2x+4\right)=42\)

\(\Rightarrow x\left(x^2-25\right)-\left(x^3+8\right)=42\)

\(\Rightarrow x^3-25x-x^3-8=42\)

\(\Rightarrow-25x-8=42\)

\(\Rightarrow-25x=50\)

\(\Rightarrow x=\dfrac{50}{-25}=-2\)

c) \(\left(x-2\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+6\left(x+1\right)^2=49\)

\(\Rightarrow x^3-6x^2+12x-8-\left(x^3-27\right)+6\left(x^2+2x+1\right)=49\)

\(\Rightarrow x^3-6x^2+12x-8-x^3+27+6x^2+12x+6=49\)

\(\Rightarrow24x+25=49\)

\(\Rightarrow24x=24\)

\(\Rightarrow x=\dfrac{24}{24}=1\)

giờ làm vẫn đc đúng ko bạn

\(\Leftrightarrow6\left(x^2-x-6\right)-3\left(x^2-4x+4\right)-3\left(x^2-1\right)=1\)

\(\Leftrightarrow6x^2-6x-36-3x^2+12x-12-3x^2+3=1\)

\(\Leftrightarrow6x=46\)

hay x=23/3