Cho A = 31+32+33+....+32006
a) Thu gon A
b) Tim x de 2A +3 +3x
DN
Những câu hỏi liên quan
Cac AC giup em bai nay voi a
Cho A=1=3=3^2+3^3+....+3^125
a,Thu gon A(Tinh A)
b,Tim chu so tan cung cua B
c,Tim x de 2A=3^2x-1
d,Chung minh rang A chia het cho 4
\(a,A=1+3+3^2+...+3^{125}\\ \Rightarrow3A=3+3^2+3^3+...+3^{126}\\ \Rightarrow2A=3^{126}-1\\ \Rightarrow A=\dfrac{3^{126}-1}{2}\\ c,2A=3^{2x}-1\\ \Rightarrow3^{126}-1=3^x-1\\ \Rightarrow x=126\)
\(d,A=\left(1+3\right)+\left(3^2+3^3\right)+...+\left(3^{124}+3^{125}\right)\\ A=\left(1+3\right)+3^2\left(1+3\right)+...+3^{124}\left(1+3\right)\\ A=\left(1+3\right)\left(1+3^2+...+3^{124}\right)\\ A=4\left(1+3^2+...+3^{124}\right)⋮4\)
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cho M=((x+2/3x)+(2/x+1)-3):(2-4x/x+1)-(3x-3x^2+1/3x)
a. rut gon M
b. tim x sao cho M<1/3
c. tim xϵZ de MϵZ
giup mk vs, mk dang can gap a
cho biểu thức A left(frac{3}{4}+frac{3}{x^2+3x}right)chialeft(frac{x^2}{27-3x}+frac{1}{x+3}right)a, rut gon A b, tim x de a-1c, tim cac gia tri nguyen cua x de A co gia tri nguyencho bthuc B left(frac{x^2}{x^3-4x}+frac{6}{6-3x}+frac{1}{x-2}right)chialeft(x-2+frac{16-x^2}{x+2}right)rut gon B tính b khi /x/ 1/2tim x de b2tim x in z de b in z
Đọc tiếp
cho biểu thức A= \(\left(\frac{3}{4}+\frac{3}{x^2+3x}\right)chia\left(\frac{x^2}{27-3x}+\frac{1}{x+3}\right)\)
a, rut gon A
b, tim x de a<-1
c, tim cac gia tri nguyen cua x de A co gia tri nguyen
cho bthuc B = \(\left(\frac{x^2}{x^3-4x}+\frac{6}{6-3x}+\frac{1}{x-2}\right)chia\left(x-2+\frac{16-x^2}{x+2}\right)\)rut gon B tính b khi /x/ = 1/2tim x de b=2tim x \(\in\) z de b \(\in\) z
Bài 2:
a: \(B=\left(\dfrac{x}{\left(x-2\right)\left(x+2\right)}-\dfrac{6}{3\left(x-2\right)}+\dfrac{1}{x-2}\right):\left(\dfrac{x^2-4+16-x^2}{x+2}\right)\)
\(=\left(\dfrac{x}{\left(x-2\right)\left(x+2\right)}-\dfrac{2}{x-2}+\dfrac{1}{x-2}\right):\dfrac{12}{x+2}\)
\(=\left(\dfrac{x}{\left(x-2\right)\left(x+2\right)}-\dfrac{1}{x-2}\right):\dfrac{12}{x+2}\)
\(=\dfrac{x-x-2}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x+2}{12}=\dfrac{-1}{6\left(x-2\right)}\)
b: Thay x=1/2 vào B, ta được:
\(B=\dfrac{-1}{6\cdot\left(\dfrac{1}{2}-2\right)}=\dfrac{-1}{6\cdot\dfrac{-3}{2}}=\dfrac{1}{9}\)
Thay x=-1/2 vào B, ta được:
\(B=\dfrac{-1}{6\cdot\left(-\dfrac{1}{2}-2\right)}=-\dfrac{1}{15}\)
c: Để B=2 thì \(\dfrac{-1}{6\left(x-2\right)}=2\)
=>6(x-2)=-1/2
=>x-2=-1/12
hay x=23/12
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A=1(x+3) - | 3x+2 |
a)rut gon A
b)tim x de A=5
Mik can rat gap giup mik di mik tick cho :(
a)th1:x+3-3x+2=(-2x)+5
th2:x+3+3x+2=4x+5
b)x=0
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Bai 1: Cho Adfrac{a^2+2a+1}{a-1}:(dfrac{a+1}{a}-dfrac{1}{1-a}+dfrac{2-a^2}{a^2-a})
a,Rut gon
b,Tinh A biet |4a-7| 1
c,Tim a de A0
d,Tim GTNN cua A
e, So sanh A voidfrac{-1}{2}
Bai 2:P(dfrac{x-1}{x+1}-dfrac{x}{x-1}-dfrac{3x+1}{1-x^2}):dfrac{2x+1}{x^2-1}
a,Rut gon
b,Tim x de Pdfrac{3}{x-1}
c,Tim gia tri nguyen cua x de P nhan gia tri nguyen
Đọc tiếp
Bai 1: Cho A=\(\dfrac{a^2+2a+1}{a-1}\):(\(\dfrac{a+1}{a}\)-\(\dfrac{1}{1-a}\)+\(\dfrac{2-a^2}{a^2-a}\))
a,Rut gon
b,Tinh A biet |4a-7| =1
c,Tim a de A>0
d,Tim GTNN cua A
e, So sanh A voi\(\dfrac{-1}{2}\)
Bai 2:P=(\(\dfrac{x-1}{x+1}\)-\(\dfrac{x}{x-1}\)-\(\dfrac{3x+1}{1-x^2}\)):\(\dfrac{2x+1}{x^2-1}\)
a,Rut gon
b,Tim x de P=\(\dfrac{3}{x-1}\)
c,Tim gia tri nguyen cua x de P nhan gia tri nguyen
Bài 2:
a, ĐKXĐ: \(x\ne\pm1;x\ne\dfrac{-1}{2}\)
\(P=\left(\dfrac{x-1}{x+1}-\dfrac{x}{x-1}-\dfrac{3x+1}{1-x^2}\right):\dfrac{2x+1}{x^2-1}\)
\(P=\left(\dfrac{x-1}{x+1}-\dfrac{x}{x-1}+\dfrac{3x+1}{x^2-1}\right).\dfrac{x^2-1}{2x+1}\)
\(P=\dfrac{\left(x-1\right)^2-x\left(x+1\right)+3x+1}{\left(x-1\right)\left(x+1\right)}.\dfrac{\left(x-1\right)\left(x+1\right)}{2x+1}\)
\(P=\dfrac{x^2-2x+1-x^2-x+3x+1}{\left(x-1\right)\left(x+1\right)}.\dfrac{\left(x-1\right)\left(x+1\right)}{2x+1}\)
\(P=\dfrac{2}{2x+1}\)
b, ĐKXĐ: \(x\ne\pm1;x\ne\dfrac{-1}{2}\)
Để \(P=\dfrac{3}{x-1}\Leftrightarrow\dfrac{2}{2x+1}=\dfrac{3}{x-1}\Leftrightarrow2\left(x-1\right)=3\left(2x+1\right)\)
\(\Leftrightarrow2x-2=6x+3\)\(\Leftrightarrow-4x=5\Leftrightarrow x=\dfrac{-5}{4}\)(TMĐK)
c, \(ĐKXĐ:x\ne\pm1;x\ne\dfrac{-1}{2}\)
Để \(P\in Z\Leftrightarrow\dfrac{2}{2x+1}\in Z\Leftrightarrow2x+1\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)
+) Với \(2x+1=1\Leftrightarrow x=0\left(TMĐK\right)\)
+) Với \(2x+1=-1\Leftrightarrow x=-1\left(KTMĐK\right)\)
+) Với \(2x+1=2\Leftrightarrow x=\dfrac{1}{2}\left(TMĐK\right)\)
+) Với \(2x+1=-2\Leftrightarrow x=\dfrac{-3}{2}\left(TMĐK\right)\)
Vậy để \(P\in Z\Leftrightarrow x\in\left\{0;\dfrac{1}{2};\dfrac{-3}{2}\right\}\)
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bai 1
a = (3x / 2x + 4 ) + (x +3 /x ^ 2 - 4 )
a . tim x de gia tri phan thuc a duoc xac dinh
b. rut gon a
c. tinh gia trin cua a khi x bang -3
d . tim gia tri cua x de phan thuc co gia tri bang 2
bai 1
a = (3x / 2x + 4 ) + (x +3 /x ^ 2 - 4 )
a . tim x de gia tri phan thuc a duoc xac dinh
b. rut gon a
c. tinh gia trin cua a khi x bang -3
d . tim gia tri cua x de phan thuc co gia tri bang 2
a: ĐKXĐ: x<>2; x<>-2
b: \(A=\dfrac{3x\left(x-2\right)+2x+6}{2\left(x-2\right)\left(x+2\right)}=\dfrac{3x^2-6x+2x+6}{2\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{3x^2+4x+6}{2\left(x-2\right)\left(x+2\right)}\)
c: Khi x=-3 thì \(A=\dfrac{3\cdot\left(-3\right)^2-4\cdot3+6}{2\left(-3-2\right)\left(-3+2\right)}=\dfrac{21}{10}\)
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cho bieu thuc A=\(\left(\dfrac{6x+4}{3\sqrt{3x^3}-8}-\dfrac{\sqrt{3x}}{3x+2\sqrt{3x}+4}\right)\left(\dfrac{1+3\sqrt{3x^3}}{1+\sqrt{3x}}-\sqrt{3x}\right)\)
a.rut gon bieu thuc A
b. tim cac gia tri nguyen x de A nguyen
a: \(A=\left(\dfrac{6x+4}{\left(\sqrt{3x}-2\right)\left(3x+2\sqrt{3x}+4\right)}-\dfrac{\sqrt{3x}}{3x+2\sqrt{3x}+4}\right)\left(\dfrac{1+\left(\sqrt{3x}\right)^3}{1+\sqrt{3x}}-\sqrt{3x}\right)\)
\(=\dfrac{6x+4-3x+2\sqrt{3x}}{\left(\sqrt{3x}-2\right)\left(3x+2\sqrt{3x}+4\right)}\cdot\left(1-\sqrt{3x}\right)^2\)
\(=\dfrac{\left(\sqrt{3x}-1\right)^2}{\sqrt{3x}-2}\)
b: Để A là số nguyên thì \(3x-2\sqrt{3x}+1⋮\sqrt{3x}-2\)
=>\(\sqrt{3x}-2\in\left\{1;-1;3;-3\right\}\)
=>\(3x\in\left\{9;1;25\right\}\)
hay x=3
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cho hai da thuc A(x)=2x(x-2)-5(x+3)+7x^3 va B(x)=-x(x+5)-(2x-3)+x(3x^2-2x).a, thu gon A(x),B(x).b, tim nghiem cua da thuc P(x)=A(x)-B(x)-x^2(4x+5)