x^4-4x^3+5x^2-2x-20

=x^4-4x^3+4x^2+x^2-2x-20

=x^2(x^2-4x+4)+x^2-2x-20

=x^2(x-2)^2 + x^2-2x+1-21

=x^2(x-2)^2+(x-1)^2-21=0

<=>x^2(x-2)^2+(x-1)^2=21

từ đây bạn giải ra cx này phải đề là tìm nghiệm nguyên nhé :D

shitbo không biết làm thì thôi ...

\(x^4-4x^3+5x^2-2x-20=0\)

\(\Leftrightarrow\left(x^2-2x\right)^2+x^2-2x-20=0\)

Đặt \(x^2-2x=a\left(a\ge-1\right)\)

\(\Rightarrow pt:a^2+a-20=0\)

\(\Leftrightarrow\left(a-4\right)\left(a+5\right)=0\)

\(\Leftrightarrow a=4\left(Do\text{ }a\ge-1\right)\)

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

\(\Leftrightarrow\left(x-1\right)^2=5\)

\(\Leftrightarrow\orbr{\begin{cases}x-1=\sqrt{5}\\x-1=-\sqrt{5}\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\sqrt{5}+1\\x=-\sqrt{5}+1\end{cases}}\)

\(\left(x+2\right)\left(x+3\right)\left(x+8\right)\left(x+12\right)-3x^2=0\)

\(\Leftrightarrow\left[\left(x+2\right)\left(x+12\right)\right]\left[\left(x+3\right)\left(x+8\right)\right]-3x^2=0\)

\(\Leftrightarrow\left(x^2+14x+24\right)\left(x^2+11x+24\right)-3x^2=0\)

Đặt \(x^2+11x+24=a\)

\(\Rightarrow pt:a\left(a+3x\right)-3x^2=0\)

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

\(\Leftrightarrow4a^2+12ax-12x^2=0\)

\(\Leftrightarrow\left(2a+3x\right)^2=21x^2\)

\(\Leftrightarrow\orbr{\begin{cases}2a+3x=x\sqrt{21}\\2a+3x=-x\sqrt{21}\end{cases}}\)

*Với \(2a+3x=x\sqrt{21}\)

\(\Leftrightarrow2x^2+22x+48+3x-x\sqrt{21}=0\)

\(\Leftrightarrow2x^2+x\left(25-\sqrt{21}\right)+48=0\)

Có \(\Delta=262-50\sqrt{21}>0\)

Nên pt có nghiệm \(x=\frac{\sqrt{21}-25\pm\sqrt{262-50\sqrt{21}}}{4}\)

Trường hợp còn lại làm tương tự

Bài 1 :

Mình nghĩ phải sửa đề ntn :

\(4\left(2x+7\right)^2-9\left(x+3\right)^2=0\)

\(\Leftrightarrow\left[2\left(2x+7\right)\right]^2-\left[3\left(x+3\right)\right]^2=0\)

\(\Leftrightarrow\left[2\left(2x+7\right)-3\left(x+3\right)\right]\left[2\left(2x+7\right)+3\left(x+3\right)\right]=0\)

\(\Leftrightarrow\left(4x+14-3x-9\right)\left(4x+14+3x+9\right)=0\)

\(\Leftrightarrow\left(x+5\right)\left(7x+23\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+5=0\\7x+23=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-5\\x=\frac{-23}{7}\end{cases}}}\)

Vậy....

b) \(A=\left(x^2+x+1\right)\left(x^2+x+2\right)-12\)

Đặt \(q=x^2+x+1\)ta có :

\(A=q\left(q+1\right)-12\)

\(A=q^2+q-12\)

\(A=q^2+4q-3q-12\)

\(A=q\left(q+4\right)-3\left(q+4\right)\)

\(A=\left(q+4\right)\left(q-3\right)\)

Thay \(q=x^2+x+1\)ta có :

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

\(A=\left(x^2+x+5\right)\left(x^2+x-2\right)\)

\(A=\left(x^2+x+5\right)\left(x^2+2x-x-2\right)\)

\(A=\left(x^2+x+5\right)\left[x\left(x+2\right)-\left(x+2\right)\right]\)

\(A=\left(x^2+x+5\right)\left(x+2\right)\left(x-1\right)\)

Cảm ơn ạ><

17. Nửa chu vi miếng đất là: \(48:2=24\left(m\right)\)

Gọi chiều rộng, chiều dài miếng đất ban đầu lần lượt là a (m) và b (m) \(\left(0< a;b< 24\right)\)

Theo bài ra, ta có:

\(\hept{\begin{cases}a+b=24\\\left(a-2\right)\left(b+6\right)-ab=12\end{cases}}\Leftrightarrow\hept{\begin{cases}a+b=24\\6a-2b=24\end{cases}}\Leftrightarrow\hept{\begin{cases}a=9\\b=15\end{cases}}\)(thỏa mãn)

Diện tích miếng đất ban đầu là: \(a.b=9.15=135\left(m^2\right)\)

--xyz=4 => √xyz=2xyz=2
--Xét:
*√zx+2√z+2=√zx+2√z+√xyz=√z(√xy+√x+2)zx+2z+2=zx+2z+xyz=z(xy+x+2)
*Tương tự suy ra √xy+√x+2=√x(√yz+√y+1)xy+x+2=x(yz+y+1)
--Thay vào ta có
*2√z√zx+2√z+2=2√xy+√x+22zzx+2z+2=2xy+x+2
*2√z√zx+2√z+2+√x√xy+√x+2=√x+2√xy+√x+2=√x+√xyz√x(√yz+√y+1)=√yz+1√yz+√y+12zzx+2z+2+xxy+x+2=x+2xy+x+2=x+xyzx(yz+y+1)=yz+1yz+y+1
--Đến đây cộng với Số hạng còn lại ta được A =1 
=>√A=1.....A=1.....
p/s: có chỗ nào sai bạn nhắc mình nha

\(\sqrt{A}=1...A=1\)  giải thích hộ mình đoạn đấy với ạ :>

\(\text{Giải}\)

\(\frac{x+2}{93}+\frac{x+3}{92}=\frac{x+4}{91}+\frac{x+5}{90}\)

\(\Leftrightarrow\frac{x+95}{93}+\frac{x+95}{92}-\frac{x+95}{91}+\frac{x+95}{90}=0\)

\(\Leftrightarrow\left(x+95\right)\left(\frac{1}{93}+\frac{1}{92}-\frac{1}{91}-\frac{1}{90}\right)=0\)

Dễ thấy thừa số thứ 2 khác 0

nên: x+95=0=>x=-95

Vậy: x=-95

cộng 2 vế với 2 tức là cộng mỗi phân số với 1.Sau đó được mâu sô chung là 95 rồi khử mẫu và làm như bình thường ,.BẠN NHÉ !

1.How far is it from your house to school ?

1.It is not very far to go to the station railway

1. It's very unusual for a military campaign to have been fought with so little loss of life.

=> Seldom has a military campaign been fought with so little loss of life.

2. It's unusal for the interior of the island to be visited by tourists.

=> Seldom is the interior of the island visited by tourists.

3. The existence of extraterrestrial is not confirmed by the report.

=> In no way is the existence of extraterrestrial life confirmed by the report.

c3; Đảo ngữ với " Not"

1. He is my friend as well as yours.

=> Not only is he your friend, but he’s mine too.

2.Burglars stole a thousand pounds' worth of electrical goods, and left the flat in an awful mess.

=> Not only did the burglars steal a thousand pounds’ worth of electrical goods, they also left the flat in an awful mess.

3. You will enhance your posture and improve your acting ability on this course.

=> Not only will you enhance your posture but you will also improve your acting ability on this course.

c4: Đảo ngữ vs " Only"

1. It wasn't until last week that the Agriculture Minister admitted defeat

=> Only until last week did the Agriculture Minister admit defeat.

2. They didn't get round to business until they had finished eating.

=> Only after they had finished eating did they get round to business.

3. They had to wait for 12 hours before their flights left.

=> Only after they had waited for twelve hours did their flight leave.

\(a)5-\left(x-6\right)=4\left(3-2x\right)\)

\(\Leftrightarrow5-x+6=12-8x\)

\(\Leftrightarrow-x+8x=12-5-6\)

\(\Leftrightarrow7x=1\Leftrightarrow x=\frac{1}{7}\)

a) 5-(x-6)=4(3-2x)

<=>5-x-6=12-8x

<=>-x+8x=2-5-6

<=>7x=1

<=>x=1/7

BO là phân giác của góc B trong tam giác ABE , nên : 

\(\frac{AB}{BE}=\frac{AO}{OE}=\frac{3}{2}\), suy ra 

\(BE=\frac{AB.2}{3}=\frac{12.2}{3}=8\left(cm\right)\)

BD là phân giác của góc B tromh tam giác ABC , nên : 

\(\frac{AB}{BC}=\frac{AD}{DC}\)= 6/7 , 

do đó   CE = 14 - 8 = 6 ( cm ) .

AE là phân giác của góc A trong tam giác ABC nên , : 

\(\frac{AC}{AB}=\frac{EC}{EB}\)= 6/8 ( 3/4 ) . Vậy AC = \(\frac{AB.3}{4}=\frac{12.3}{4}=9\left(cm\right)\)