-(x+11)=(3x +22) - (6+x)
-(-x+111-134)=199-(x-24)
(x+3).2+x=(200-9x)-350
Ét o ét ;((
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a: P(x)=5x^2-4x+7
Sửa đề: Q(x)=-5x^3-x^2+4x-5
Q(x)+P(x)+5x^2-2=0
=>5x^2-4x+7-5x^3-x^2+4x-5+5x^2-2=0
=>5x^3=0
=>x=0
\(a)P\left(x\right)=5x^5+3x-4x^4-2x^3+6+4x^2\)
\(P\left(x\right)=5x^5-4x^4-2x^3+4x^2+3x+6\)
\(Q\left(x\right)=2x^4-x+3x^2-2x^3+\dfrac{1}{4}-x^5\)
\(Q\left(x\right)=-x^5+2x^4-2x^3+3x^2-x+\dfrac{1}{4}\)
\(a)P\left(x\right)-Q\left(x\right)=\left(5x^5-4x^4-2x^3+4x^2+3x+6\right)+\left(-x^5+2x^4-2x^3+3x^2-x+\dfrac{1}{4}\right)\)
\(P\left(x\right)-Q\left(x\right)=5x^5-4x^4-2x^3+4x^2+3x+6+x^5-2x^4+2x^3-3x^2+x-\dfrac{1}{4}\)
\(P\left(x\right)-Q\left(x\right)=\left(5x^5+x^5\right)+\left(-4x^4-2x^4\right)+\left(-2x^3+2x^3\right)+\left(4x^2-3x^2\right)+\left(3x+x\right)+\left(6-\dfrac{1}{4}\right)\)
\(P\left(x\right)-Q\left(x\right)=6x^5-6x^4+x^2+4x+\dfrac{23}{4}\)
\(\text{c)Thay x=-1 vào biểu thức P(x),ta được:}\)
\(P\left(x\right)=5.\left(-1\right)^5-4.\left(-1\right)^4-2.\left(-1\right)^3+4.\left(-1\right)^2+3.\left(-1\right)+6\)
\(P\left(x\right)=\left(-5\right)-4-\left(-2\right)+4+\left(-3\right)+6\)
\(P\left(x\right)=\left(-9\right)-\left(-2\right)+4+\left(-3\right)+6\)
\(P\left(x\right)=\left(-7\right)+4+\left(-3\right)+6\)
\(P\left(x\right)=\left(-3\right)+\left(-3\right)+6\)
\(P\left(x\right)=\left(-6\right)+6=0\)
\(\text{Vậy giá trị của P(x) tại x=-1 là:0}\)
\(\text{Vậy =-1 là nghiệm của P(x)}\)
\(\text{Thay x=-1 vào biểu thức Q(x),ta được:}\)
\(Q\left(x\right)=\left(-1\right).5+2.\left(-1\right)^4-2.\left(-1\right)^3+3.\left(-1\right)^2-\left(-1\right)+\dfrac{1}{4}\)
\(Q\left(x\right)=\left(-5\right)+2-\left(-2\right)+3-\left(-1\right)+\dfrac{1}{4}\)
\(Q\left(x\right)=\left(-3\right)-\left(-2\right)+3-\left(-1\right)+\dfrac{1}{4}\)
\(Q\left(x\right)=\left(-5\right)+3-\left(-1\right)+\dfrac{1}{4}\)
\(Q\left(x\right)=\left(-2\right)-\left(-1\right)+\dfrac{1}{4}\)
\(Q\left(x\right)=\left(-3\right)+\dfrac{1}{4}=\dfrac{-13}{4}\)
\(\text{Vậy x=-1 không phải là nghiệm của Q(x)}\)
\(\text{d)Thay x=-1 vào biểu thức }P\left(x\right)-Q\left(x\right),\text{ta được:}\)
\(P\left(x\right)-Q\left(x\right)=6.\left(-1\right)^5-6.\left(-1\right)^4+\left(-1\right)^2+4.\left(-1\right)+\dfrac{23}{4}\)
\(P\left(x\right)-Q\left(x\right)=\left(-6\right)-6+1+\left(-4\right)+\dfrac{23}{4}\)
\(P\left(x\right)-Q\left(x\right)=\left(-12\right)+1+\left(-4\right)+\dfrac{23}{4}\)
\(P\left(x\right)-Q\left(x\right)=\left(-11\right)+\left(-4\right)+\dfrac{23}{4}\)
\(P\left(x\right)-Q\left(x\right)=\left(-15\right)+\dfrac{23}{4}=\dfrac{-37}{4}\)
\(\text{Vậy giá trị của P(x)-Q(x) tại x=-1 là:}\dfrac{-37}{4}\)
\(4,7\div0,25+5,3\times4\)
\(=18,8+21,2\)
\(=40\)
\(3\times\left(a-2\right)+150=240\)
\(3\times\left(a-2\right)=90\)
\(a-2=30\)
\(a=32\)
\(\dfrac{1}{9}+a+\dfrac{7}{12}=\dfrac{17}{18}\)
\(\dfrac{1}{9}+a=\dfrac{13}{36}\)
\(a=\dfrac{1}{4}\)
\(\left(\dfrac{1}{2}\times\dfrac{1}{3}+\dfrac{1}{3}\times\dfrac{1}{4}+\dfrac{1}{4}\times\dfrac{1}{5}+\dfrac{1}{5}\times\dfrac{1}{6}+\dfrac{1}{6}\times\dfrac{1}{7}+\dfrac{1}{7}\times\dfrac{1}{8}\right)\times a=\dfrac{9}{16}\)
\(\left(\dfrac{1}{2\times3}+\dfrac{1}{3\times4}+\dfrac{1}{4\times5}+\dfrac{1}{5\times6}+\dfrac{1}{6\times7}+\dfrac{1}{7\times8}\right)\times a=\dfrac{9}{16}\)
\(\left(\dfrac{1}{2}-\dfrac{1}{8}\right)\times a=\dfrac{9}{16}\)
\(\dfrac{3}{8}\times a=\dfrac{9}{16}\)
\(a=\dfrac{3}{2}\)
1) 42/35 - [ 33/14 . 21/22 - ( 54/25 : 27/25 - 7/10 )]
= 42/35 - ( 9/4-13/10)
=42/35-19/20
=1/4
2) a) 5/21 . x/34 = 35/102
x/34=35/102 : 5/21
x/34=49/34
=> x=34
Vậy x=34
b) 15/x . 21/13 = 45/91
15/x=45/91 : 21/13
15/x=15/49
=> x=49
Vậy x=49
\(\left(1-x\right)^2=2003.\left(x-1\right)\)
\(\left(1-x\right)^2-2003\left(x-1\right)=0\)
\(\left(1-x\right)^2+2003\left(1-x\right)=0\)
\(\left(1-x\right)\left(1-x+2003\right)=0\)
\(\left(1-x\right)\left(2004-x\right)=0\)
\(TH1:1-x=0\)
\(x=1\)
\(TH2:2004-x=0\)
\(x=2004\)
vậy........
b: =>3x+9=0 và y^2-9=0 và x+y=0
=>x=-3; y=3
a: (2x-5)(y+3)=-22
mà x,y là số nguyên
nên \(\left(2x-5;y+3\right)\in\left\{\left(1;-22\right);\left(11;-2\right);\left(-1;22\right);\left(-11;2\right)\right\}\)
=>\(\left(x,y\right)\in\left\{\left(3;-25\right);\left(8;-5\right);\left(2;19\right);\left(-3;-1\right)\right\}\)
hông biết☹☹
\(-\left(x+11\right)=\left(3x+22\right)-\left(6+x\right)\)
\(\Rightarrow-x-11=3x+22-6-x\)
\(\Rightarrow-x-11=2x+16\)
\(\Rightarrow-x-2x=11+16\)
\(\Rightarrow-3x=27\)
\(\Rightarrow x=-9\)
\(-\left(-x+11-134\right)=199-\left(x-24\right)\)
\(\Rightarrow x-111+134=199-x+24\)
\(\Rightarrow x+23=233-x\)
\(\Rightarrow2x=200\)
\(\Rightarrow x=100\)
\(\left(x+3\right).2+x=\left(200-9x\right)-350\)
\(\Rightarrow2x+6+x=200-9x-350\)
\(\Rightarrow12x=-156\)
\(\Rightarrow x=-13\)