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

Tính giá trị biểu thức

ý d, e, f

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

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

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

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

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

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

\(A=x^{100}-12x^{99}+12x^{98}-12x^{97}+...-12x^3+12x^2-12x+12\)

Thay x = 11 ta có:

\(A=11^{100}-12.11^{99}+12.11^{98}-...-12.11^3+12.11^2-12.11+12\)

\(=11^{100}-12\left(11^{99}-11^{98}+11^{97}-...+11^3-11^2+11\right)+12\)

Đặt \(B=11^{99}-11^{98}+...+11\)

\(\Rightarrow11B=11^{100}-11^{99}+...+11^2\)

\(\Rightarrow12B=11^{100}+11\)

\(\Rightarrow B=\dfrac{11^{100}+11}{12}\)

Từ đó, \(A=11^{100}-12.\dfrac{11^{100}+11}{12}+12\)

\(=11^{100}-11^{100}-11+12=1\)

Vậy A = 1

Ta có: \(x=11\Rightarrow x+1=12\)

Khi đó, ta được:

\(A=x^{100}-12x^{99}+12x^{98}-12x^{97}+...-12x^3+12x^2-12x+12\)

\(=x^{100}-\left(x+1\right)x^{99}+\left(x+1\right)x^{98}-\left(x+1\right)x^{97}+...-\left(x+1\right)x^3+\left(x+1\right)x^2-\left(x+1\right)x+12\)

\(=x^{100}-x^{100}-x^{99}+x^{99}+x^{98}-x^{98}-x^{97}+...-x^4-x^3+x^3+x^2-x^2-x+12\)

\(=\left(x^{100}-x^{100}\right)-\left(x^{99}-x^{99}\right)+\left(x^{98}-x^{98}\right)-...-\left(x^3-x^3\right)+\left(x^2-x^2\right)-x+12\)

\(=0-x+12=0-11+12=-11+12=1\)

Vậy tại x=11 thì A=1

tại x=11 nhé

x4−12x3+12x2−12x+111 tại x=11

có thể phân tích thành nhân tử để tính nha bạn

a) Ta có P = ( 4 x 2 − 1 ) ( 2 x + 1 ) − ( 2 x − 1 ) − ( 4 x 2 − 1 ) ( 2 x + 1 ) ( 2 x − 1 ) = 3 − 4 x 2  

b) Ta có  Q = 3 x ( x + 3 ) . ( x + 3 ) ( x − 3 ) − x = 9 − 3 x x + 3

a) Ta có: \(\left(x-1\right)^2+2=x^2+3x\)

\(\Leftrightarrow x^2-2x+1+2-x^2-3x=0\)

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

hay \(x=\dfrac{3}{5}\)

câu b sai đề bn

câu c tìm gì

a: A=-x^3+3x^2-3x+1

=-x^3+1+3x^2-3x

=-(x-1)(x^2+x+1)+3x(x-1)

=(x-1)(-x^2-x-1+3x)

=(x-1)(-x^2+2x-1)

=-(x-1)^3

Khi x=6 thì A=-(6-1)^3=-125

b: B=8-12x+6x^2-x^3

=2^3-3*2^2*x+3*2*x^2-x^3

=(2-x)^3

Khi x=12 thì B=(2-12)^3=(-10)^3=-1000

\(a,-x^3+3x^2-3x+1\)

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

\(=-\left(x-1\right)^3\)

Thay \(x=6\) vào biểu thức trên, ta được:
\(-\left(6-1\right)^3\)

\(=-5^3\)

\(=-125\)

\(b,8-12x+6x^2-x^3\)

\(=2^3-3\cdot2^2\cdot x+3\cdot2\cdot x^2-x^3\)

\(=\left(2-x\right)^3\)

Thay \(x=12\) vào biểu thức trên, ta được:

\(\left(2-12\right)^3\)

\(=\left(-10\right)^3\)

\(=-1000\)

#\(Toru\)

Đề lỗi rồi. Bạn xem lại

b: =>x(8-7)=-33

=>x=-33

c: =>-12x+60+21-7x=5

=>-19x=-76

hay x=4

d: =>-2x-2-x+5+2x=0

=>3-x=0

hay x=3

a) \(-12\left(x-5\right)+7\left(3-x\right)=5\)

\(-12x+60+21-7x=5\)

\(-19x=5-60-21\)

\(-19x=-76\)

\(x=4\)

vậy \(x=4\)

tương tự