\(-4+x=-4+7\)

\(\Rightarrow x=-4+7+4\)

\(\Rightarrow x=7\)

\(\left(-3\right)-\left(-5-x\right)=-15+\left(-6+15\right)\)

\(\Rightarrow x=-15-6+15+3-5\)

\(\Rightarrow x=-8\)

\(-\left(x+2\right)+2=\left(8-15\right)+15\)

\(\Rightarrow-x+2+2=-7+15\)

\(\Rightarrow-x=-7+15-2-2\)

\(\Rightarrow x=-4\)

\(4+\left(x-4\right)=-\left(11-2\right)+\left(11+4\right)\)

\(\Rightarrow4+x-4=-11+2+11+4\)

\(\Rightarrow x=-11+2+11+4-4+4\)

\(\Rightarrow x=6\)

a) -4 + x = -4 + 7            b) (-3) - (-5-x) = -15 + (-6+15)

   -4 + x + 4 - 7 = 0            (-3) + 5 + x = -6

        x - 7 = 0                     2 + 6 = -x

        x= 7                            8 = -x 

 Vậy x= 7                            x= -8   =) Vậy x= -8

c) -(x+2)+2=(8-15)+15          d) 4+(x-4)= -(11-2) + (11+4 )

   -x - 2 + 2 = 8                       4+x-4 = 6

   -x = 8                                     x=6

    x= -8                              Vậy x=6

Vậy x= -8

a

\(x^2\left(2x+15\right)+4\left(2x+15\right)=0\\ \Leftrightarrow\left(2x+15\right)\left(x^2+4\right)=0\\ \Leftrightarrow2x+15=0\left(x^2+4>0\forall x\right)\\ \Leftrightarrow2x=-15\\ \Leftrightarrow x=-\dfrac{15}{2}\)

b

\(5x\left(x-2\right)-3\left(x-2\right)=0\\ \Leftrightarrow\left(x-2\right)\left(5x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-2=0\\5x-3=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0+2=2\\x=\dfrac{0+3}{5}=\dfrac{3}{5}\end{matrix}\right.\)

c

\(2\left(x+3\right)-x^2-3x=0\\ \Leftrightarrow2\left(x+3\right)-\left(x^2+3x\right)=0\\ \Leftrightarrow2\left(x+3\right)-x\left(x+3\right)=0\\ \Leftrightarrow\left(x+3\right)\left(2-x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x+3=0\\2-x=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0-3=-3\\x=2-0=2\end{matrix}\right.\)

a: =>(2x+15)(x^2+4)=0

=>2x+15=0

=>2x=-15

=>x=-15/2

b; =>(x-2)(5x-3)=0

=>x=2 hoặc x=3/5

c: =>(x+3)(2-x)=0

=>x=2 hoặc x=-3

Mình trả lời hai câu đầu trước nha!

1)

x= -6+2

x= -4

2)

=-5 x = 13 + (-3)

= -5 .x =10

x = 10 ÷ -5

x = -2

Tìm max B biết B=15−4x−x2B=15-4x-x^2B=15−4x−x2

DELL THỂ hiểu đc đề ghi đề như shi* vậy :(

x2 là J tui chưa thấy đề nào như zầy cả :)))

a. 2(x-1) + (x+2) - (x+3) = 15 - (x+1)

=>2x-2+x+2-x-3=15-x-1

=>(2x+x-x)-2+2-3=15-1-x

=>2x-3=14-x

=>3x=17

=>x=17/3

b. x+1/15 + x+2/14 = x+4/12 + x+5/11

\(\Rightarrow\frac{x+1}{15}+1+\frac{x+2}{14}+1=\frac{x+4}{12}+1+\frac{x+5}{11}+1\)

\(\Rightarrow\frac{x+16}{15}+\frac{x+16}{14}=\frac{x+16}{12}+\frac{x+16}{11}\)

\(\Rightarrow\frac{x+16}{15}+\frac{x+16}{14}-\frac{x+16}{12}-\frac{x+16}{11}=0\)

\(\Rightarrow\left(x+16\right)\left(\frac{1}{15}+\frac{1}{14}-\frac{1}{12}-\frac{1}{11}\right)=0\)

\(\Rightarrow x+16=0\).Do \(\frac{1}{15}+\frac{1}{14}-\frac{1}{12}-\frac{1}{11}\ne0\)

=>x=-16

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:

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

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

\(\left(x+2\right)\left(x^2-2x+4\right)-x\left(x^2-2\right)=15\)

\(\Leftrightarrow x^3+8-x^3+2x=15\)

\(\Leftrightarrow2x+8=15\)

\(\Leftrightarrow2x=15-8=7\)

\(\Leftrightarrow x=\frac{7}{2}\)

\(\left(x+2\right)\left(x^2-2x+4\right)-x.\left(x^2-2\right)=15\)

a: Ta có: \(\left(x+2\right)\left(x^2-2x+4\right)-x\left(x^2+2\right)=15\)

\(\Leftrightarrow x^3+8-x^3-2x=15\)

\(\Leftrightarrow2x=-7\)

hay \(x=-\dfrac{7}{2}\)

b: Ta có: \(\left(x-2\right)^3-\left(x-4\right)\left(x^2+4x+16\right)+6\left(x+1\right)^2=49\)

\(\Leftrightarrow x^3-6x^2+12x-8-x^3+64+6\left(x+1\right)^2=49\)

\(\Leftrightarrow-6x^2+12x+56+6x^2+12x+6=49\)

\(\Leftrightarrow24x=-13\)

hay \(x=-\dfrac{13}{24}\)