\(d,=24x^2-38x+3\\ e,=x^2-12x+35\\ f,=\left(x^2-144\right)\left(4x-1\right)=4x^3-x^2-576x+144\)

Tính giá trị biểu thức

ý d, e, f

d: \(\left(12x-1\right)\left(2x-3\right)\)

\(=24x^2-36x-2x+3\)

\(=24x^2-38x+3\)

\(12\left(x+5\right)+2x=130\\\Leftrightarrow 12x+60+2x=130\\ \Leftrightarrow14x=70\\ \Leftrightarrow x=5\\ ----\\ 23\left(x-5\right)-12x=138\\ \Leftrightarrow23x-115-12x=138\\ \Leftrightarrow23x-12x=138+115\\ \Leftrightarrow11x=253\\ \Leftrightarrow x=\dfrac{253}{11}=23\\ ----\\ 360-12x+23\left(x-5\right)=278\\ \Leftrightarrow360-12x+23x-115=278\\ \Leftrightarrow-12x+23x=278+115-360\\ \Leftrightarrow11x=33\\ \Leftrightarrow x=\dfrac{33}{11}=3\)

\(6\left(x+3\right)+3\left(x-5\right)=278\\ \Leftrightarrow6x+18-3x-15=278\\ \Leftrightarrow6x-3x=278+15-18\\ \Leftrightarrow3x=275\\ \Leftrightarrow x=\dfrac{275}{3}\\ ---\\ \left(7-x\right)\left(3x-90\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}7-x=0\\3x-90=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=7\\x=30\end{matrix}\right.\)

\(\left(x-2\right)\left(4x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-2=0\\4x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\4x=2\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{1}{2}\end{matrix}\right.\\ ---\\ \left(2x-18\right)\left(3x-9\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}2x-18=0\\3x-9=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=18\\3x=9\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=9\\x=3\end{matrix}\right.\)

Câu 9:

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

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

\(9,\Leftrightarrow x^2\left(x-2\right)-\left(x-2\right)=0\\ \Leftrightarrow\left(x-2\right)\left(x-1\right)\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\\x=2\end{matrix}\right.\\ 11,\Leftrightarrow x^2+5x-x-5=0\\ \Leftrightarrow\left(x+5\right)\left(x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-5\end{matrix}\right.\\ 12,\Leftrightarrow\left(x+1\right)^2-36=0\\ \Leftrightarrow\left(x+7\right)\left(x-5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-7\\x=5\end{matrix}\right.\\ 13,\Leftrightarrow x^3-25x-x^3-8=17\\ \Leftrightarrow-25x=25\Leftrightarrow x=-1\\ 14,\Leftrightarrow x\left(2x^2+8x-3x-12\right)=0\\ \Leftrightarrow x\left(x+4\right)\left(2x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-4\\x=\dfrac{3}{2}\end{matrix}\right.\)

\(9,x^3-2x^2-x+2=0\\ \Rightarrow x^2\left(x-2\right)-\left(x-2\right)=0\\ \Rightarrow\left(x^2-1\right)\left(x-2\right)=0\\ \Rightarrow\left(x-1\right)\left(x+1\right)\left(x-2\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=1\\x=-1\\x=2\end{matrix}\right.\)

\(10,\) giống 9

\(11,x^2+4x-5=0\\ \Rightarrow\left(x^2-x\right)+\left(5x-5\right)=0\\ \Rightarrow x\left(x-1\right)+5\left(x-1\right)=0\\ \Rightarrow\left(x-1\right)\left(x+5\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=1\\x=-5\end{matrix}\right.\)

\(12,2x^2+4x+2=72\\ \Rightarrow2x^2+4x-70=0\\ \Rightarrow x^2+2x-35=0\\ \Rightarrow\left(x^2-5x\right)+\left(7x-35\right)=0\\ \Rightarrow x\left(x-5\right)+7\left(x-5\right)=0\\ \Rightarrow\left(x-5\right)\left(x+7\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=5\\x=-7\end{matrix}\right.\)

\(13,x\left(x-5\right)\left(x+5\right)-\left(x+2\right)\left(x^2-2x+4\right)=17\\ \Rightarrow x\left(x^2-25\right)-\left(x^3+8\right)=17\\ \Rightarrow x^3-25x-x^3-8=17\\ \Rightarrow-25x=25\\ \Rightarrow x=-1\)

\(14,2x^3+5x^2-12x=0\\ \Rightarrow x\left(2x^2+5x-12\right)=0\\ \Rightarrow x\left[\left(2x^2+8x\right)-\left(3x+12\right)\right]=0\\ \Rightarrow x\left[2x\left(x+4\right)-3\left(x+4\right)\right]=0\\ \Rightarrow x\left(2x-3\right)\left(x+4\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{3}{2}\\x=-4\end{matrix}\right.\)

a) x - (11 - x) = -48 + (-12 + x)

x - 11 + x = -48 + (-12) + x

x + x - x = -48 + (-12) + 11

x + 0 = (-60) + 11

x = -49

b) (15 - x) + (x - 12) = 7 - (-8 + x)

15 - x + x - 12 = 7 + 8 + x

15 - x + x - 12 = 15 + x

x + x - x = 15 - 15 + 12

x + 0 = 0 + 12

x = 12

c) (x - 12) - (2x + 31) = -6 - 5

x - 12 - 2x - 31 = -1

x - 12 - 2x = -1 + 31

x - 12 - 2x = 30

x - 2x = 30 + 12

-1x = 42

=> x = -42

d) |x + 5| - (-17) = 20

=> |x + 5| + 17 = 20

=> |x + 5| = 20 - 17

=> |x + 5| = 3

=> \(\orbr{\begin{cases}x+5=3\\x+5=-3\end{cases}}\)

=> \(\orbr{\begin{cases}x=3-5\\x=-3-5\end{cases}}\)

=> \(\orbr{\begin{cases}x=-2\\x=-8\end{cases}}\)

12x² - 3x = 0

=> 12x² - 3x = (2² . 3x²) - 3x = 0

=> 3x . (4x - 1) = 0

=> 3x . 4x - 3x . 1 = 0

=> 3x . 4x - 3x = 0

=> 12x - 3x = 0

=> 4x  =0

=> x = 0

x=0 chac chan

Thanks bạn nhìu!

a, 3x.(12x-4)-9x(4x-3)=30

=>36x²-12x-36x²+27x=30

=>5x=30

=> x=6

b,x.(5-2x)+2x.(x-1)=15

=> 5x-2x²+2x²-2x=15

=>3x=15

=>x=5

tk mk nha bn

*****Chúc bạn học giỏi*****

a) 3x . (12x - 4) - 9x(4x - 3) = 30

 3x . 12x - 12x - 9x.4x + 27x = 30

(3x . 12x - 9x . 4x) - (12x - 27x) = 30

(36x² -36x²) + 15x = 30

=> 15x = 30

=> x = 30 : 15

=> x = 2

b) x.(5 - 2x) + 2x.(x - 1) = 15

5x - 2x² + 2x² - 2x = 15

(5x - 2x) - (2x² - 2x²) = 15

=> 3x = 15

=> x = 15 : 3 = 5

nhân đơn thức vs đa thức rồi tính thôi

 P(x) = 2x³ – 5x² + 8x – 3

          Nghiệm hữu tỷ nếu có của đa thức P(x)  trên là:

                    (– 1); 1; (–1/2); 1/2 ; (–3/2); 3/2 ; –3…

          Sau khi kiểm tra ta thấy x = 1/2 là nghiệm nên đa thức chứa nhân tử            ( x – 1/2) ­ hay (2x – 1). Do đó ta tìm cách tách các hạng tử của đa thức để xuất hiện nhân tử chung (2x – 1).

          2x³  - 5x² + 8x – 3 = 2x³- x² – 4x² + 2x + 6x – 3

                                      = x²( 2x – 1) – 2x( 2x – 1) + 3(2x – 1)

                                      = ( 2x – 1)(x² – 2x + 3).

          Hoặc chia P(x) cho (x – 1) ta được thương đúng là: x² – 2x + 3

          P(x) =  2x³  – 5x² + 8x – 3 = ( 2x – 1)(x² – 2x + 3)

     Vậy P(x) =  2x³  – 5x² + 8x – 3 = ( 2x – 1)(x² – 2x + 3)