1)15,54 : 4,2=3,7

2)45,8x4,58 + 45,8x5,42 + 4,58x100 

=45,8\(\times\)(58+42)+4,58\(\times\)100

=45,8\(\times\)100+4,58\(\times\)100

=4580+458

=5038    

47,25 -13,14 - 23,86

=47,25-(13,14+23,86)

=47,25-47,00

=0,25               
 38,24 x 15,2 + 15,2 x 61,76 

=15,2x(38,24+61,76)

=15,2x100,00

=1520

x=13 nên x+1=14

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

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

=x-1

=13-1=12

TL

T i k cho mik ik rồi mik Trả lời cho

#Kirito

Kb nha ,mik sẽ trả lời giúp bạn 

Bài này là bài ktra hk 2 của mình 

Bài 1:

\(f\left(x\right)=x^2+8x+25\)

Cho \(f\left(x\right)=0\Rightarrow x^2+8x+25=0\)

\(\Rightarrow x^2+8x+16+9=0\)

\(\Rightarrow\left(x+4\right)^2+9=0\)

Dễ thấy: \(\left(x+4\right)^2\ge0\forall x\)

\(\Rightarrow\left(x+4\right)^2+9\ge9>0\forall x\) ( vô nghiệm )

Vậy đa thức \(f\left(x\right)=x^2+8x+25\) không có nghiệm

Bài 2:

\(f\left(x\right)=x^{14}-14x^{13}+14x^{12}-...+14x^2-14x+14\)

\(f\left(x\right)=x^{14}-\left(13+1\right)x^{13}+\left(13+1\right)x^{12}-...+\left(13+1\right)x^2-\left(13+1\right)x+\left(13+1\right)\)

Do \(f\left(x\right)=13\) nên ta chỗ nào có \(13\) ta thay bằng \(x\)

\(f\left(13\right)=x^{14}-\left(x+1\right)x^{13}+\left(x+1\right)x^{12}-...+\left(x+1\right)x^2-\left(x+1\right)x+\left(x+1\right)\)

\(f\left(13\right)=x^{14}-x^{14}-x^3+x^{13}+x^{12}-...+x^3+x^2-x^2-x+x+1=1\)

Vậy \(f\left(13\right)=1\)

a:Sửa đề: \(\sqrt{x+\sqrt{14x-49}}+\sqrt{x-\sqrt{14x-49}}=\sqrt{14}\)

ĐKXĐ: x>=7/2

\(\sqrt{x+\sqrt{14x-49}}+\sqrt{x-\sqrt{14x-49}}=\sqrt{14}\)

=>\(\sqrt{14x+14\cdot\sqrt{14x-49}}+\sqrt{14x-14\sqrt{14x-49}}=14\)

=>\(\sqrt{14x-49+2\cdot\sqrt{14x-49}\cdot7+49}+\sqrt{14x-49-2\cdot\sqrt{14x-49}\cdot7+49}=14\)

=>\(\sqrt{\left(\sqrt{14x-49}-7\right)^2}+\sqrt{\left(\sqrt{14x-49}+7\right)^2}=14\)

=>\(\left|\sqrt{14x-49}-7\right|+\left|\sqrt{14x-49}+7\right|=14\)

=>\(\left|\sqrt{14x-49}-7\right|=14-\sqrt{14x-49}-7=7-\sqrt{14x-49}\)

=>\(\sqrt{14x-49}-7\le0\)

=>\(\sqrt{14x-49}\le7\)

=>14x-49<=49

=>14x<=98

=>x<=7

=>7/2<=x<=7

b: ĐKXĐ:x>=1

Sửa đề: \(\sqrt{x+2\sqrt{x-1}}-\sqrt{x-2\sqrt{x-1}}=1\)

=>\(\sqrt{x-1+2\cdot\sqrt{x-1}\cdot1+1}-\sqrt{x-1-2\cdot\sqrt{x-1}\cdot1+1}=1\)

=>\(\sqrt{\left(\sqrt{x-1}+1\right)^2}-\sqrt{\left(\sqrt{x-1}-1\right)^2}=1\)

=>\(\sqrt{x-1}+1-\left|\sqrt{x-1}-1\right|=1\)

=>\(\left|\sqrt{x-1}-1\right|=\sqrt{x-1}\)

=>\(\left[\begin{array}{l}\sqrt{x-1}-1=\sqrt{x-1}\\ \sqrt{x-1}-1=-\sqrt{x-1}\end{array}\right.\Rightarrow\left[\begin{array}{l}-1=0\left(loại\right)\\ 2\sqrt{x-1}=1\end{array}\right.\)

=>\(\sqrt{x-1}=\frac12\)

=>x-1=1/4

=>x=5/4(nhận)