\(B=10-5x\left(x-1,2\right)+2x\left(2,5x-3\right)\)

\(=10-5x^2+6x+5x^2-6x\)

\(=10\)

Vì 10 là hằng số => giá trị của B không phụ thuộc vào biến x

\(10-5x.\left(x-1,2\right)+2x.\left(2,5x-3\right)\)

\(=10-\left(5x^2-6x\right)+\left(5x^2-6x\right)\)

\(=10-5x^2+6x+5x^2-6x\)

\(=10+\left(-5x^2+5x^2\right)+\left(6x-6x\right)\)

\(=10\)

Ta có: \(B=10-5x\left(x-1,2\right)+2x\left(2,5x-3\right)\)

\(=10-5x^2+6x+5x^2-6x\)

=10

Vậy: Giá trị của biểu thức \(B=10-5x\left(x-1,2\right)+2x\left(2,5x-3\right)\) luôn luôn không đổi

1)\(\left(4x-10\right)\left(24+5x\right)=0\)

\(\Leftrightarrow2\left(2x-5\right)\left(24+5x\right)=0\)

Vì 2≠0

nên \(\left[{}\begin{matrix}2x-5=0\\24+5x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=5\\5x=-24\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{5}{2}\\x=\frac{-24}{5}\end{matrix}\right.\)

Vậy: \(x\in\left\{\frac{5}{2};\frac{-24}{5}\right\}\)

2) \(0,5x\left(x-3\right)=\left(x-3\right)\left(2,5x-4\right)\)

\(\Leftrightarrow0,5x\left(x-3\right)-\left(x-3\right)\left(2,5x-4\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left[0,5x-\left(2,5x-4\right)\right]=0\)

\(\Leftrightarrow\left(x-3\right)\left(0,5x-2,5x+4\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(-2x+4\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(4-2x\right)=0\)

\(\Leftrightarrow\left(x-3\right)\cdot2\cdot\left(2-x\right)=0\)

Vì 2≠0

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

Vậy: x∈{2;3}

3) \(4x^2-1=\left(2x+1\right)\left(3x-5\right)\)

\(\Leftrightarrow\left(2x+1\right)\left(2x-1\right)-\left(2x+1\right)\left(3x-5\right)=0\)

\(\Leftrightarrow\left(2x+1\right)\left[2x-1-\left(3x-5\right)\right]=0\)

\(\Leftrightarrow\left(2x+1\right)\left(2x-1-3x+5\right)=0\)

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

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

Vậy: \(x\in\left\{\frac{-1}{2};4\right\}\)

4) \(\left(2-3x\right)\left(x+11\right)=\left(3x-2\right)\left(2-5x\right)\)

\(\Leftrightarrow\left(2-3x\right)\left(x+11\right)-\left(3x-2\right)\left(2-5x\right)=0\)

\(\Leftrightarrow\left(2-3x\right)\left(x+11\right)+\left(2-3x\right)\left(2-5x\right)=0\)

\(\Leftrightarrow\left(2-3x\right)\left(x+11+2-5x\right)=0\)

\(\Leftrightarrow\left(2-3x\right)\left(13-4x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2-3x=0\\13-4x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=2\\4x=13\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{2}{3}\\x=\frac{13}{4}\end{matrix}\right.\)

Vậy: \(x\in\left\{\frac{2}{3};\frac{13}{4}\right\}\)

dễ ***** ra

vler

a/ \(5x^3-2x^2+4x-4\)

x \(x^3+3x^2-5x-1\)

  \(5x^6-6x^4-20x^2+4\)

b/ \(-4,2x^4+3,1x^2-\)\(\)7

 x       \(2,5x^3-\)7x + 1,5

\(-10,5x^{ }\)7\(-21,\)7\(x^3-10,5\)

\(a.-\frac{5}{9}x+1=\frac{2}{3}x-10\\ \Leftrightarrow-\frac{5}{9}x-\frac{2}{3}x=-1-10\\\Leftrightarrow -\frac{11}{9}x=-11\\ \Leftrightarrow x=9\)

Vậy nghiệm của phương trình trên là \(9\)

\(b.3-5x=6x-2\\\Leftrightarrow -5x-6x=-3-2\\\Leftrightarrow -11x=-5\\ \Leftrightarrow x=\frac{5}{11}\)

Vậy nghiệm của phương trình trên là \(\frac{5}{11}\)

\(c.3\left(x-1\right)=5+3x\\ \Leftrightarrow3x-3=5+3x\\ \Leftrightarrow3x-3x=3+5\\\Leftrightarrow 0x=8\)

\(\Rightarrow\) Vô nghiệm

\(d.2\left(1-2,5x\right)+5x=0\\ \Leftrightarrow2-5x+5x=0\\ \Leftrightarrow2=0\left(sai\right)\)

\(\Rightarrow\) Vô nghĩa (Vô nghiệm)

a: \(\left|-2,5x\right|=x-12\)

=>\(2,5\left|x\right|=x-12\)

=>\(\begin{cases}x-12\ge0\\ \left(2,5x\right)^2=\left(x-12\right)^2\end{cases}\Rightarrow\begin{cases}x\ge12\\ \left(2,5x-x+12\right)\left(2,5x+x-12\right)=0\end{cases}\)

=>\(\begin{cases}x\ge12\\ \left(1,5x+12\right)\left(3,5x-12\right)=0\end{cases}\Rightarrow\begin{cases}x\ge12\\ x\in\left\lbrace-8;\frac{24}{7}\right\rbrace\end{cases}\)

=>x∈∅

b: |5x|-3x-2=0(1)

TH1: x>=0

=>5x>=0

(1) sẽ trở thành:

5x-3x-2=0

=>2x-2=0

=>2x=2

=>x=1(nhận)

TH2: x<0

=>5x<0

(1) sẽ trở thành:

-5x-3x-2=0

=>-8x=2

=>x=-1/4(nhận)

c: \(\left|-2x\right|+x-5x-3=0\)

=>2|x|-4x-3=0(2)

TH1: x>=0

(2) sẽ trở thành: 2x-4x-3=0

=>-2x-3=0

=>2x=-3

=>x=-3/2(loại)

TH2: x<0

(2) sẽ trở thành: -2x-4x-3=0

=>2x+4x+3=0

=>6x=-3

=>x=-1/2(nhận)

d: \(\left|3-x\right|+x^2-x\left(x+4\right)=0\)

=>\(\left|x-3\right|+x^2-x^2-4x=0\)

=>|x-3|=4x

=>\(\begin{cases}4x\ge0\\ \left(4x\right)^2=\left(x-3\right)^2\end{cases}\Rightarrow\begin{cases}x\ge0\\ \left(4x-x+3\right)\left(4x+x-3\right)=0\end{cases}\)

=>\(\begin{cases}x\ge0\\ \left(3x+3\right)\left(5x-3\right)=0\end{cases}\)

=>x=3/5

e: \(\left(x-1\right)^2+\left|x+21\right|-x^2-13=0\)

=>\(x^2-2x+1+\left|x+21\right|-x^2-13=0\)

=>-2x-12+|x+21|=0

=>|x+21|=2x+12

=>\(\begin{cases}2x+12\ge0\\ \left(2x+12\right)^2=\left(x+21\right)^2\end{cases}\Rightarrow\begin{cases}x\ge-6\\ \left(2x+12-x-21\right)\left(2x+12+x+21\right)=0\end{cases}\)

=>\(\begin{cases}x\ge-6\\ \left(x-9\right)\left(3x+33\right)=0\end{cases}\Rightarrow x=9\)

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

TH1 : \(4+2x\ge0\Leftrightarrow2x\ge-4\Leftrightarrow x\ge-2\)

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

\(\Leftrightarrow2x+4x=-4\)

\(\Leftrightarrow6x=-4\)

\(\Leftrightarrow x=-\dfrac{2}{3}\) (t/m)

TH2 : \(4+2x< 0\Leftrightarrow2x< -4\Leftrightarrow x< -2\)

\(\text{- (4 + 2x) = -4x}\)

\(\Leftrightarrow-4-2x=-4x\)

\(\Leftrightarrow-2x+4x=4\)

\(\Leftrightarrow2x=4\)

\(\Leftrightarrow x=2\) (ko t/m)

\(S=\left\{-\dfrac{2}{3}\right\}\)

b) \(\left|-2,5x\right|=x-12\)

TH1 : \(-2,5x\ge0\Leftrightarrow x\le0\)

\(-2,5x=x-12\)

\(\Leftrightarrow-2,5x-x=-12\)

\(\Leftrightarrow-3,5x=-12\)

\(\Leftrightarrow x=\dfrac{24}{7}\) (ko t/m)

TH2 : \(-2,5x< 0\Leftrightarrow x>0\)

\(\text{2,5x = x - 12}\)

\(\Leftrightarrow2,5x-x=-12\)

\(\Leftrightarrow1,5x=-12\)

\(\Leftrightarrow x=-8\) (ko t/m)

\(S=\varnothing\)

c) \(\left|-2x\right|+x-5x-3=0\)

\(\Leftrightarrow\left|-2x\right|-4x-3=0\)

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

TH1 : \(-2x\ge0\Leftrightarrow x\le0\)

\(-2x=3+4x\)

\(\Leftrightarrow-2x-4x=3\)

\(\Leftrightarrow-6x=3\)

\(\Leftrightarrow x=-\dfrac{1}{2}\) (t/m)

TH2 : \(-2x< 0\Leftrightarrow x>0\)

\(\text{2x = 3 + 4x}\)

\(\Leftrightarrow2x-4x=3\)

\(\Leftrightarrow-2x=3\)

\(\Leftrightarrow x=-\dfrac{3}{2}\) (ko t/m)

\(S=\left\{-\dfrac{1}{2}\right\}\)