a, \(\left\{{}\begin{matrix}x=1\\y=1\end{matrix}\right.\)

b,\(\left\{{}\begin{matrix}x=3\\y=4\end{matrix}\right.\)

c,\(\left\{{}\begin{matrix}x=1\\y=1\end{matrix}\right.\)

d,\(\left\{{}\begin{matrix}x=0\\y=2\end{matrix}\right.\)

\(x^2+x+1=y^2\\ \Leftrightarrow\left(x+\dfrac{1}{2}\right)^2+1=y^2\\ \Leftrightarrow\left(x+\dfrac{1}{2}\right)^2-y^2=-1\\ \Leftrightarrow\left(x-y+\dfrac{1}{2}\right)\left(x+y+\dfrac{1}{2}\right)=-1=\left(-1\right)\cdot1\\ TH_1:\left\{{}\begin{matrix}x-y+\dfrac{1}{2}=-1\\x+y+\dfrac{1}{2}=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-y=-\dfrac{3}{2}\\x+y=\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{1}{2}\\y=1\end{matrix}\right.\)

\(TH_2:\left\{{}\begin{matrix}x-y+\dfrac{1}{2}=1\\x+y+\dfrac{1}{2}=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-y=\dfrac{1}{2}\\x+y=-\dfrac{3}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{1}{2}\\y=-1\end{matrix}\right.\)

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

\(\left\{{}\begin{matrix}150x-135y=15\\150x+210y=360\end{matrix}\right.\)

\(\Leftrightarrow-345y=-345\)

\(\Rightarrow y=1\left(1\right)\)

Thay (1) vào ptr đầu: \(10x-9\cdot1=1\)

\(\Rightarrow y=1\)

a, Ta có : \(\left\{{}\begin{matrix}3x+2y=-2\\-x+4y=3\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}3\left(4y-3\right)+2y=-2\\x=4y-3\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}12y-9+2y=-2\\x=4y-3\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}14y=7\\x=4y-3\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}y=\frac{1}{2}\\x=\frac{4.1}{2}-3=-1\end{matrix}\right.\)

Vậy hệ phương trình có duy nhất 1 nghiệm là \(\left(x;y\right)=\left(-1;\frac{1}{2}\right)\)

b, Ta có : \(\left\{{}\begin{matrix}x+2y=11\\5x-3y=3\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}x=11-2y\\5\left(11-2y\right)-3y=3\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}x=11-2y\\55-10y-3y=3\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}x=11-2y\\-13y=-52\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}x=11-2.4=3\\y=4\end{matrix}\right.\)

Vậy hệ phương trình có duy nhất 1 nghiệm là \(\left(x;y\right)=\left(3;4\right)\)

c, Ta có : \(\left\{{}\begin{matrix}10x-9y=1\\15x+21y=36\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}30x-27y=3\\30x+42y=72\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}10x-9y=1\\-69y=-69\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}10x-9=1\\y=1\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}x=1\\y=1\end{matrix}\right.\)

Vậy hệ phương trình có duy nhất 1 nghiệm là \(\left(x;y\right)=\left(1;1\right)\)

d, Ta có : \(\left\{{}\begin{matrix}2x+y=3\\x+y=2\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}y=3-2x\\x+2-2x=2\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}y=3-2x\\2-x=2\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}y=3-2.0=3\\x=0\end{matrix}\right.\)

Vậy hệ phương trình có duy nhất 1 nghiệm là \(\left(x;y\right)=\left(0;3\right)\)

e, Ta có : \(\left\{{}\begin{matrix}x+y=2\\2x-3y=9\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}x=2-y\\2\left(2-y\right)-3y=9\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}x=2-y\\4-2y-3y=9\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}x=2-y\\-5y=5\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}x=2+1=3\\y=-1\end{matrix}\right.\)

Vậy hệ phương trình có duy nhất 1 nghiệm là \(\left(x;y\right)=\left(3;-1\right)\)

f, Ta có : \(\left\{{}\begin{matrix}x-2y=11\\5x+3y=3\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}x=11+2y\\5\left(11+2y\right)+3y=3\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}x=11+2y\\55+10y+3y=3\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}x=11+2y\\13y=-52\end{matrix}\right.\)​

Vậy hệ phương trình có duy nhất 1 nghiệm là \(\left(x;y\right)=\left(3;-4\right)\)

g, Ta có : \(\left\{{}\begin{matrix}3x-y=5\\2x+3y=18\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}y=3x-5\\2x+3\left(3x-5\right)=18\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}y=3x-5\\2x+9x-15=18\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}y=3x-5\\11x=33\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}y=9-5=4\\x=3\end{matrix}\right.\)

Vậy hệ phương trình có duy nhất 1 nghiệm là \(\left(x;y\right)=\left(3;4\right)\)

h, Ta có : \(\left\{{}\begin{matrix}5x+3y=-7\\3x-y=-8\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}5x+3\left(3x+8\right)=-7\\y=3x+8\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}5x+9x+24=-7\\y=3x+8\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}14x=-31\\y=3x+8\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}x=-\frac{31}{14}\\y=3.\left(-\frac{31}{14}\right)+8=\frac{19}{14}\end{matrix}\right.\)

Vậy hệ phương trình có duy nhất 1 nghiệm là \(\left(x;y\right)=\left(-\frac{31}{14};\frac{19}{14}\right)\)

...????

a) \(y=\dfrac{5-3x}{2}\)

b) \(y=\dfrac{5x-6}{3}\)

c) \(y=\dfrac{4x-5}{3}\)

d) \(y=\dfrac{7-3x}{2}\)

a. 3x + 2y = 5

<=> 2(1,5x + y) = 5

<=> 1,5x + y = 2,5

<=> \(\left[{}\begin{matrix}x=\dfrac{2,5-y}{1,5}\\y=2,5-1,5x\end{matrix}\right.\) 

các câu còn lại tương tự

a) ta có : \(N=-21x^{99}-21x^{98}-...-21x^2-21x\)

\(\Rightarrow xN=-21x^{100}-21x^{99}-...-21x^2-21x^2\)

\(\Rightarrow xN-N=-21x^{100}+21x\)

\(\Leftrightarrow\left(x-1\right)N=-21x^{100}+21x\Leftrightarrow N=\dfrac{21x-21x^{100}}{x-1}\)

\(\Rightarrow A=x^{100}-21x^{99}-21x^{98}-...-21x^2-21x+2010\)

\(=x^{100}+\dfrac{21x-21x^{100}}{x-1}+2010\)

\(=\dfrac{21x-21x^{100}+x^{101}-x^{100}+2010x-2010}{x-1}\)

\(=\dfrac{x^{101}-22x^{100}+2031x-2010}{x-1}\)

thay \(x=22\) ta có : \(A=\dfrac{22^{101}-22.22^{100}+2031.22-2010}{22-1}\)

\(=\dfrac{22^{101}-22^{101}+2031.22-2010}{21}=\dfrac{2031.22-2010}{21}=2032\)

vậy ............................................................................................................

câu b lm tương tự .