0

0

Ta có :

\(\left|15x-7\right|-\left|15x+7\right|=0\)

\(\Rightarrow\left|15x-7\right|=\left|15x+7\right|\)

=> 15x-7=15x+7 hoặc 15x-7 =-(15x+7)

Xét : 15x-7=15x+7

=> 15x=15x+14

=>0=14 (vô lý )

Xét : 15x-7=-(15x+7)

=> 15x-7=-15x-7

=> 15x=-15x

=> 15x -(-15x)=0

=> 30x=0 => x=0

Vậy x=0

tick mik nha !!

7/5 x X + 3/8 x X= 9

X x (7/5 +3/8)=9

X x 71/40 = 9

X             = 9:71/40

X             = 360/71

     \(\frac{7}{15}X+\frac{3}{8}X=9\)

     \(\left(\frac{7}{15}+\frac{3}{8}\right)X=9\) 

\(\left(\frac{56}{120}+\frac{45}{120}\right)X=9\)

                  \(\frac{101}{120}X=9\)

                          \(X=9:\frac{101}{120}\)

                          \(X=9x\frac{120}{101}\)

                          \(X=\frac{1080}{101}\)

\(15x-13x+5x=49:7\)

\(\Leftrightarrow7x=7\Leftrightarrow x=1\)

\(15x-13x+5x=49:7\)

\(\Leftrightarrow2x+5x=7\)

\(\Leftrightarrow7x=7\)

\(\Leftrightarrow x=1\)

\(15x-13x+5x=49:7\)

\(\Leftrightarrow7x=7\)

\(\Leftrightarrow x=1\)

\(A=x^{10}-14x^9-x^9+14x^8+x^8-14x^7-x^7...-x+14+1\)

\(A=x^9\left(x-14\right)-x^8\left(x-14\right)+x^7\left(x-14\right)-...-x\left(x-14\right)+1\)

\(A=1\) (Do x=14)

C(x)= 2x-3=0 hoac 5x+7=0

        2x=0+3        5x=0-7

        2x=3            5x=-7

         x=3:2            x=-7:5

          x=1.5            x=-1.4

a.

\(\left(2x-3\right)\times\left(5x+7\right)=0\)

TH1:

\(2x-3=0\)

\(2x=3\)

\(x=\frac{3}{2}\)

TH2:

\(5x+7=0\)

\(5x=-7\)

\(x=-\frac{7}{5}\)

Vậy \(C\left(x\right)\) có nghiệm là \(\frac{3}{2}\) hoặc \(-\frac{7}{5}\)

b.

\(\left(15x^5+4x^2-8\right)-\left(15x^5-x-8\right)=0\)

\(15x^5+4x^2-8-15x^5+x+8=0\)

\(\left(15x^5-15x^5\right)+4x^2+x+\left(8-8\right)=0\)

\(x\left(4x-1\right)=0\)

TH1:

\(x=0\)

TH2:

\(4x-1=0\)

\(4x=1\)

\(x=\frac{1}{4}\)

Vậy \(D\left(x\right)\) có nghiệm là \(0\) hoặc \(\frac{1}{4}\)

c.

\(\left(5x^7-8x^2\right)-\left(4x^7+4^2\right)-\left(x^7+4\right)=0\)

\(5x^7-8x^2-4x^7-16-x^7-4=0\)

\(\left(5x^7-4x^7-x^7\right)-8x^2-\left(16-4\right)=0\)

\(-8x^2-12=0\)

\(-8x^2=12\)

\(x^2=-\frac{12}{8}\)

mà \(x^2\ge0\) với mọi x

=> \(E\left(x\right)\) vô nghiệm

\(a,C\left(x\right)=\left(2x-3\right)\left(5x+7\right)=0\)

\(\Leftrightarrow\) \(\left[\begin{array}{nghiempt}2x-3=0\\5x+7=0\end{array}\right.\) \(\Leftrightarrow\) \(\left[\begin{array}{nghiempt}x=\frac{3}{2}\\x=-\frac{7}{5}\end{array}\right.\)

Vậy \(x=\frac{3}{2}\) và \(x=-\frac{7}{5}\) là nghiệm của đa thức C(x)

\(b,D\left(x\right)=\left(15x^5+4x^2-8\right)-\left(15x^5-x-8\right)=0\)

\(\Leftrightarrow15x^5+4x^2-8-15x^5+x+8=0\)

\(\Leftrightarrow4x^2+x=0\) \(\Leftrightarrow x\left(4x+1\right)=0\)  \(\Leftrightarrow\) \(\left[\begin{array}{nghiempt}x=0\\4x+1=0\end{array}\right.\)  \(\Leftrightarrow\) \(\left[\begin{array}{nghiempt}x=0\\x=-\frac{1}{4}\end{array}\right.\)

Vậy \(x=0\) và \(x=-\frac{1}{4}\) là nghiệm đa thức D(x)

\(c,E\left(x\right)=\left(5x^7-8x^2\right)-\left(4x^7+4x^4\right)-\left(x^7+4\right)=0\)

\(\Leftrightarrow5x^7-8x^2-4x^7-4x^4-x^7-4=0\)

\(\Leftrightarrow-8x^2-4x^4-4=0\)

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

\(\Leftrightarrow2x^2+x^4+1=0\) \(\Leftrightarrow x^4+x^2+x^2+1=0\) 

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

\(\Leftrightarrow\left(x^2+1\right)^2=0\) \(\Leftrightarrow x^2+1=0\) \(\Leftrightarrow x^2=-1\) \(\Rightarrow x\in\varnothing\)

Vậy E(x) vô nghiệm

Ta có x=14 suy ra x+1=15

Do đó thay x+1 vào H(x), ta được:

H(14)= x¹⁰ - (x+1)x⁹ +(x+1)x⁸-(x+1)x⁷+...+ (x+1)x² - (x+1)x + x+1

H(14)= x^10 - x^10 -x^9 +x^8- x^8-x^7 +....+ x^3 +x^2 -x^2-x+x +1

Hay H(14)=1

Đặt Q(x) = x⁹ - x⁸ + x⁷ - ... + x - 1 thì (x + 1) * Q(x) = (x + 1) * (x⁹ - x⁸ + x⁷ - ... + x - 1) = x¹⁰ - 1 \(\Rightarrow\left(14+1\right)\cdot Q\left(14\right)=14^{10}-1\)

Dễ thấy: H(x) = x¹⁰ - 15* Q(x) \(\Rightarrow H\left(14\right)=14^{10}-\left(14^{10}-1\right)=1\)

Xin cậu !

3( x - 5 )( x - 2 )( x + 2 ) + 4 = 7 + 3x³ - 15x²

<=> ( 3x - 15 )( x² - 4 ) + 4 - 7 = 3x³ - 15x²

<=> 3x³ - 12x - 15x² + 60 - 3 = 3x³ - 15x²

<=> 57 = 3x³ - 15x² - 3x³ + 12x + 15x²

<=> 57 = 12x

<=> x = 57/12 = 19/4

Tìm x biết:

3(x - 5)(x - 2)(x + 2) + 4 = 7 + 3x³ - 15x²

\(3\left(x-5\right)\left(x-2\right)\left(x+2\right)+4=3x^3-15x^2-12x=64\)

\(7+3^3+\left(-15\right)x^2=3x^3-15x^2+7\)

\(3x^3-15x^2-12x+64=3x^3-15x^2+7\)

\(\Rightarrow\frac{19}{4}\)

\(3\left(x-5\right)\left(x-2\right)\left(x+2\right)+4=7+3x^3-15x^2\)

\(\Leftrightarrow3\left(x-5\right)\left(x^2-4\right)+4=7+3x^3-15x^2\)

\(\Leftrightarrow3\left(x^3-4x-5x^2+20\right)+4=7+3x^3-15x^2\)

\(\Leftrightarrow3x^3-12x-15x^2+60+4=7+3x^3-15x^2\)

\(\Leftrightarrow-12x+64=7\Leftrightarrow-12x=-57\Leftrightarrow x=\frac{19}{4}\)