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a)\(A=x^5-36x^4+37x^3-69x^2+34x+15\)
=\(x^5-35x^4-x^4+35x^3+2x^2-70x^2+x^2-35x+x+15\)
=\(\left(x^4-x^3+x^2+x\right)\left(x-35\right)+x+15\)
=0+35+15=50(do x=35)
b, \(3\left(6x-5\right)\left(4x+1\right)-\left(8x+3\right)\left(9x-2\right)=203\)
\(\Rightarrow3\left(24x^2+6x-20x-5\right)-\left(72x^2-16x+27x-6\right)=203\)
\(\Rightarrow72x^2-42x-15-72x^2-11x+6=203\)
\(\Rightarrow-53x=203-6+15=212\)
\(\Rightarrow x=-4\)
Chúc bạn học tốt!!!
Bài 2:
b: \(\left(2x+3\right)\left(x-4\right)+\left(x-5\right)\left(x-2\right)=\left(3x-5\right)\left(x-4\right)\)
\(\Leftrightarrow2x^2-8x+3x-12+x^2-7x+10=3x^2-12x-5x+20\)
\(\Leftrightarrow3x^2-12x-2=3x^2-17x+20\)
=>-12x-2=-17x+20
=>5x=22
hay x=22/5
c: \(\left(8x-3\right)\left(3x+2\right)-\left(4x+7\right)\left(x+4\right)=\left(2x+1\right)\left(5x-1\right)\)
\(\Leftrightarrow24x^2+16x-9x-6-\left(4x^2+16x+7x+28\right)=10x^2-2x+5x-1\)
\(\Leftrightarrow24x^2+7x-6-4x^2-23x-28=10x^2+3x-1\)
\(\Leftrightarrow20x^2-16x-34=10x^2+3x-1\)
\(\Leftrightarrow10x^2-19x-33=0\)
\(\text{Δ}=\left(-19\right)^2-4\cdot10\cdot\left(-33\right)=1681>0\)
Do đó: Phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{19-41}{20}=\dfrac{-22}{20}=\dfrac{-11}{10}\\x_2=\dfrac{19+41}{20}=3\end{matrix}\right.\)
Bài 2:
b)\((2x+3)(x-4)+(x-5)(x-2)=(3x-5)(x-4)\)
\(\Leftrightarrow2x^2-5x-12+x^2-7x+10=3x^2-17x+20\)
\(\Leftrightarrow3x^2-12x-2=3x^2-17x+20\)
\(\Leftrightarrow5x=22\Rightarrow x=\frac{22}{5}\)
c)\((8x-3)(3x+2)-(4x+7)(x+4)=(2x+1)(5x-1)\)
\(\Leftrightarrow24x^2+7x-6-4x^2-23x-28=10x^2+3x-1\)
\(\Leftrightarrow20x^2-16x-34=10x^2+3x-1\)
\(\Leftrightarrow10x^2-19x-33=0\)
\(\Leftrightarrow\left(x-3\right)\left(10x+11\right)=0\)
Suy ra x=3;x=-11/10
a ) Nếu \(x=71\) \(\Rightarrow70=x-1\)
Thay \(70=x-1\) vào A , ta được :
\(A=x^5-\left(x-1\right)x^4-\left(x-1\right)x^3-\left(x-1\right)x^2-\left(x-1\right)x\)
\(=x^5-x^5+x^4-x^4+x^3-x^3+x^2-x^2+x\)
\(=x\)
\(=71\)
Vậy \(A=71\) tại \(x=71\)
b ) Ta có : \(x=35\)
\(\Rightarrow\left\{{}\begin{matrix}36=x+1\\37=x+2\\69=2x-1\\34=x-1\end{matrix}\right.\) ( * )
Thay ( * ) vào B , ta được :
\(B=x^5-\left(x+1\right)x^4+\left(x+2\right)x^3-\left(2x-1\right)x^2-\left(x-1\right)x+15\)
\(=x^5-x^5-x^4+x^4+2x^3-2x^3+x^2-x^2+x+15\)
\(=x+15\)
\(=35+15=50\)
Vậy \(B=50\) tại \(x=35\)
Bài 2:
a: \(x^2\left(x^2-16\right)=0\)
\(\Leftrightarrow x\left(x-4\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\\x=-4\end{matrix}\right.\)
b: \(x^8+36x^4=0\)
\(\Leftrightarrow x^4=0\)
hay x=0
a(b+3)-b(3+b)
=(3+b)(a-b)
Thay số, có: (3+1997).(2003-1997)
= 2000.6 =12000
xy(x+y)-2x-2y
xy(x+y)- 2(x+y)
(x+y).(xy-2)
Thay số, co: 7. (8-2)
7.4=28
Ta có : \(\frac{x-35}{21}+\frac{x-36}{20}>\frac{x-37}{19}+\frac{x-38}{18}\)(1)
\(\Leftrightarrow\left(\frac{x-35}{21}-1\right)+\left(\frac{x-36}{20}-1\right)\)\(-\left(\frac{x-37}{19}-1\right)-\left(\frac{x-38}{18}-1\right)\)\(>0\)
\(\Leftrightarrow\frac{x-56}{21}+\frac{x-56}{20}-\frac{x-56}{19}-\frac{x-56}{18}\)\(>0\)
\(\Leftrightarrow\left(x-56\right)\left(\frac{1}{21}+\frac{1}{20}-\frac{1}{19}-\frac{1}{18}\right)\)\(>0\)
Vì \(\frac{1}{21}+\frac{1}{20}-\frac{1}{19}-\frac{1}{18}< 0\)
\(\Rightarrow x-56< 0\)\(\Leftrightarrow x< 56\)
Vậy tập nghiệm của BPT(1) là \(S=\left\{x\in R|x< 56\right\}\)
1/
a, \(4x^4+1=4x^4+4x^2+1-4x^2=\left(2x^2+1\right)^2-\left(2x\right)^2=\left(2x^2+2x+1\right)\left(2x^2-2x+1\right)\)
b, \(4x^4+y^4=4x^4+4x^2y^2+y^4-4x^2y^2=\left(2x^2+y^2\right)^2-\left(2xy\right)^2=\left(2x^2+2xy+y^2\right)\left(2x^2-2xy+y^2\right)\)
c, \(x^4+324=x^4+36x^2+324-36x^2=\left(x^2+18\right)^2-\left(6x\right)^2=\left(x^2+6x+18\right)\left(x^2-6x+18\right)\)
2/
a, \(x^2+\frac{1}{3}x+\frac{1}{36}=\left(x+\frac{1}{6}\right)^2=\left(\frac{35}{6}+\frac{1}{6}\right)^2=6^2=36\)
b, \(x^2-y^2+2y-1=x^2-\left(y-1\right)^2=\left(x+y-1\right)\left(x-y+1\right)=\left(100+1-1\right)\left(100-1+1\right)=100.100=10000\)
Thay x = 35/6 vào biểu thức trên ta có :
\(\left(\frac{35}{6}\right)^2+\frac{1}{3}.\frac{35}{6}+\frac{1}{36}=\frac{1225}{36}+\frac{35}{18}+\frac{1}{36}=36\)
Thay x = 100 ; y = 1 vào biểu thúc trên ta có :
\(100^2-1^2+2.1-2=10000-1+2-2=9999\)
Thay \(x=\frac{35}{6}\)vào biểu thức trên ta có :
\(\left(\frac{35}{6}\right)^2+\frac{1}{3}\cdot\frac{35}{6}+\frac{1}{36}\)
\(=\frac{1225}{36}+\frac{35}{18}+\frac{1}{36}=\frac{1225}{36}+\frac{70}{36}+\frac{1}{36}=\frac{1296}{36}=36\)
Thay x = 100,y = 1 vào biểu thức trên ta có :
1002 - 12 + 2.1 -1 = 1002 - 1 + 2 - 1 = 1002 - 1 + 1 = 1002 = 10000
Ta có:
\(x=35\)
\(\Rightarrow\left\{{}\begin{matrix}x+1=36\\x+2=27\\2x-1=69\\x-1=34\end{matrix}\right.\) (1)
Thay (1) vào biểu thức ta được:
\(x^5-\left(x+1\right)x^4+\left(x+2\right)x^3-\left(2x-1\right)x^2-\left(x-1\right)x+15\)
\(=x^5-x^5-x^4+x^4+2x^3-2x^3+x^2-x^2+x+15\)
\(=x+15\)
\(=35+15\)
\(=50\)
50