a) \(3x-\frac{3}{2}-5x+\frac{10}{3}=1\)

\(3x-5x=1+\frac{3}{2}-\frac{10}{3}\)

\(-2x=-\frac{5}{6}\)

\(x=\frac{5}{12}\)

b) \(\left|x-1\right|=5-x\)

Th1:

\(x-1=5-x\)

\(x+x=5+1\)

\(2x=6\)

\(x=3\)

Th2:

\(-\left(x-1\right)=5-x\)

\(x+1=5-x\)

\(x+x=5-1\)

\(2x=4\)

\(x=2\)

Vậy \(x=3\)và \(x=2\)

a) 3x - 5x = 1 + 3/2 - 10/3

-2x = -5/6

x = -5/6 : ( - 2 )

x = 5/12

b)  |x-1|= 5-x

Nếu x \(\ge\)1 \(\Rightarrow\)x - 1 \(\ge\)0 \(\Rightarrow\)x - 1 = 5 - x.

2x = 6

x = 3.

Nếu x < 1 \(\Rightarrow\)x - 1 < 0 \(\Rightarrow\)Ix-1I = 1 - x

\(\Rightarrow\)1 - x = 5 - x \(\Rightarrow\)vô lý.

Vậy x = 3

\(\sqrt{19+8\sqrt{3}}-\sqrt{19-8\sqrt{3}}\)

\(=\sqrt{4^2+8\sqrt{3}+\left(\sqrt{3}\right)^2}-\sqrt{4^2-8\sqrt{3}+\left(\sqrt{3}\right)^2}\)

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

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

\(=\sqrt{3}+4-\sqrt{3}+4\)

\(=8\)

\(\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}\)

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

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

\(=\left|\sqrt{x-1}+1\right|+\left|\sqrt{x-1}-1\right|\)

ở dưới làm r

3 . ( 5x + 15 ) + x - 11 = 98

=>3(5x+15)+x=109

=>15x+45+x=109

=>16x=64

=>x=4

x³ - 2x² + 6x = 12

x³ - 2x² + 6x - 12 = 0

x²(x - 2) + 6(x - 2)=0

(x - 2)(x² + 6) = 0

\(\Leftrightarrow \begin{bmatrix} x - 2 = 0 & & \\ x^{2} + 6 = 0& & \end{bmatrix}\) bỏ dấu ngoặc bên phải nha pn

\(\Leftrightarrow \begin{bmatrix} x = 2 & & \\ x^{2} = - 6 & & \end{bmatrix}\) không tìm được giá trị của x (pn ghi cái này kế pn chỗ x² = - 6 nhé

Vậy x = 2

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

\(\Rightarrow\) \(x^3-2x^2+6x-12=0\)

\(\Rightarrow x^2\left(x-2\right)+6\left(x-2\right)=0\)

\(\Rightarrow\left(x-2\right)\left(x^2+6\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x-2=0\\x^2+6=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\x\in\varnothing\end{matrix}\right.\)

Vậy $x=2$

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

\(\Rightarrow x^3-2x^2+6x-12=0\)

\(\Rightarrow x^2\left(x-2\right)+6\left(x-2\right)-0\)

\(\Rightarrow\left(x^2+6\right)\left(x-2\right)=0\)

\(\Rightarrow\left\{{}\begin{matrix}x^2+6=0\\x-2=0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x\in\varnothing\\x=2\end{matrix}\right.\)

Vậy \(x=2\)

\(3,\)Áp dụng bđt Mincopski \(\sqrt{a^2+b^2}+\sqrt{c^2+d^2}\ge\sqrt{\left(a+c\right)^2+\left(b+d\right)^2}\)hai lần có

\(VT\ge\sqrt{\left(\sqrt{x}+\sqrt{y}\right)^2+\left(\sqrt{yz}+\sqrt{zx}\right)^2}+\sqrt{z+xy}\)

       \(\ge\sqrt{\left(\sqrt{x}+\sqrt{y}+\sqrt{z}\right)^2+\left(\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\right)^2}\)

       \(=\sqrt{x+y+z+2\left(\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\right)+\left(\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\right)^2}\)

       \(=\sqrt{1+2t+t^2}\left(t=\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\right)\)
        \(=\sqrt{\left(t+1\right)^2}=t+1=VP\left(Đpcm\right)\)

\(2,\frac{2\sqrt{ab}}{\sqrt{a}+\sqrt{b}}\le\frac{2\sqrt{ab}}{2\sqrt{\sqrt{a}.\sqrt{b}}}=\sqrt{\sqrt{ab}}\left(đpcm\right)\)

<-6>.[3^2-<-2.^5]-<-11>

=<-6>.[9-<-10>]-<-11>

=<-6>.19-<-11>

=   -114-<-11>

=    -103

75  .  <-29>+29.<-25>

=<-75>.29+29.<-25>

=29.<-75+-25>

=29.<-100>

=-2900

 (x + 1) + (x + 2) + (x + 3 )+ ... + (x + 20) = 250 ( có 20 nhóm )
=> ( x + x + x +...+ x ) + ( 1 + 2 + 3 +...+ 20) = 250 ( có 20 x và 20 số hạng )
=> x . 20 + 20 . 21 : 2 = 250
=> x . 20 + 210 = 250
=> x . 20 = 250 - 210
=> x . 20 = 40
=> x = 40 : 20
     x = 2

(x+1)+(x+2)+(x+3)+...+(x+20)=250

Dãy trên có 20 số hạng

20x+(1+2+3+...+20)=250

Tổng dãy 1+2+3+...+20 là

(20+1).20:2=210

20x+210=250

20x=250-210

20x=40

x=40:20

x=2

(x + 1) + (x + 2) + (x + 3 )+ ... + (x + 20) = 250

250 = ( 1 + 2 + 3 + ... + 20 ) + ( 20 x x )

250 = 210 + ( 20 x X ) 

X = ( 250 - 210 ) : 20

X = 2. k mình nha

1) (x+1)+(x+2)+(x+3)+...+(x+100) = 5750

   (x+x+x+...+x)+(1+2+3+...+100)= 5750

   100x         +  5050                   =  5750

    100x                                      =  5750 - 5050

     100 x                                       =   700

          x                                        =  700 :100

          x                                         = 7

2) 96 : 8= 12

Bài 1

(x+1)+(x+2)+(x+3)+....+(x+100)=5750

x+1+x+2+x+3+...+x+100=5750

(x+x+x+...+x)+(1+2+3+...+100)=5750 (có 100 số x)

100x+(100.101:2)=5750

100x+5050=5750

100x=5750-5050

100x=700

x=7

Bài 2

96:8=12

PT <=> \(\frac{4x}{3}=\frac{14x}{3}+5\)

<=> \(\frac{10x}{3}=-5\)

<=> x= \(-5:\frac{10}{3}=-\frac{15}{10}=-\frac{3}{2}\)

vậy x= - 3/2