Tìm biểu thức Q, biết rằng :
\(\dfrac{x^2+2x}{x-1}.Q=\dfrac{x^2-4}{x^2-x}\)
Những câu hỏi liên quan
tim bieu thuc Q biet rang
\(\dfrac{x^2+2x}{x-1}.Q=\dfrac{x^2-4}{x^2-x}\)
\(\dfrac{x^2+2x}{x-1}\). Q = \(\dfrac{x^2-4}{x^2-x}\)
\(\Leftrightarrow\) Q = \(\dfrac{x^2-4}{x^2-x}\): \(\dfrac{x^2+2x}{x-1}\)
\(\Leftrightarrow\) Q = \(\dfrac{\left(x-2\right)\left(x+2\right)}{x\left(x-1\right)}\) . \(\dfrac{x-1}{x\left(x+2\right)}\)
\(\Leftrightarrow\) Q = \(\dfrac{x-2}{x}\)
Vậy Q = \(\dfrac{x-2}{x}\)
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Cho bieu thuc: \(Q=\left(\dfrac{x^2-2x}{2x^2+8}+\dfrac{2x^2}{x^2.\left(x-2\right)}\right).\left(\dfrac{x^2-x-2}{x^2}\right)\)
a, Rut gon bieu thuc Q
b, Tim gia tri ca x de Q co gia tri bang \(\dfrac{1}{4}\)
1)Tim he so a cua da thuc A(x)=-7x2-3y2 +9xy-2x2+y2,biet rang da thuc co1 nghiem bang\(\dfrac{1}{2}\)
2) Tim m, biet rang da thuc Q(x)=mx2 + 2mx -3co 1 nghiemx=-1
Bài 2:
Q(-1)=0
=>m-2m-3=0
=>-m-3=0
hay m=-3
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cho bieu thuc
P=\(\left(\dfrac{\sqrt{x}}{x-4}+\dfrac{1}{\sqrt{x}-2}\right).\dfrac{\sqrt{x}-2}{2}\)với x>=0,x≠4
a. tim gia tri cua P khi x=64
b. rút gọn bieu thuc p
c. tim cac gia tri cua x de bieu thuc 2P nhan gia tri nguyen
b \(P=\dfrac{\sqrt{x}+\sqrt{x}+2}{x-4}\cdot\dfrac{\sqrt{x}-2}{2}=\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\)
a: Khi x=64 thì \(P=\dfrac{8+1}{8+2}=\dfrac{9}{10}\)
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cho bieu thuc
P=\(\left(\dfrac{\sqrt{x}}{x-4}+\dfrac{1}{\sqrt{x}-2}\right).\dfrac{\sqrt{x}-2}{2}\)với x>=0,x≠4
a. tim gia tri cua P khi x=64
b. rút gọn bieu thuc p
c. tim cac gia tri cua x de bieu thuc 2P nhan gia tri nguyen
b: \(P=\dfrac{\sqrt{x}+\sqrt{x}+2}{x-4}\cdot\dfrac{\sqrt{x}-2}{2}=\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\)
a: Khi x=64 thì \(P=\dfrac{8+1}{8+2}=\dfrac{9}{10}\)
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Cho 2 bieu thuc :
A=\(\dfrac{x-3}{x+2}va\) B= \(\dfrac{3}{x+3}+\dfrac{2}{x-3}-\dfrac{3x-9}{x^2-9}\left(x-2,x\ne3x\ne-3\right)\)
a, Tinh gia tri bieu thuc A khi x=5
b, Chung minh : B=\(\dfrac{2}{x-3}\)
c, Biet C = A.B, Tim x de c = \(\dfrac{-1}{3}\)
\(a,A=\dfrac{5-3}{5+2}=\dfrac{2}{7}\\ b,B=\dfrac{3x-9+2x+6-3x+9}{\left(x-3\right)\left(x+3\right)}=\dfrac{2\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{2}{x-3}\\ c,C=AB=\dfrac{x-3}{x+2}\cdot\dfrac{2}{x-3}=\dfrac{2}{x+2}\\ C=-\dfrac{1}{3}\Leftrightarrow x+2=-6\Leftrightarrow x=-8\left(tm\right)\)
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Bai 1: Cho bt
P=(\(\dfrac{x-2}{x^2-1}\)-\(\dfrac{x+2}{x^2+2x+1}\)).(\(\dfrac{1-x^2}{2}\))2
a, Rut gon P
b, Tim x de \(\dfrac{P-4}{5}\)=x
Bai 2:Chung minh bieu thuc A=\(\dfrac{3}{2}\)-\(\dfrac{x}{x^2+x+1}\) luon duong voi moi x
Bài 1:
a: \(P=\left(\dfrac{x-2}{\left(x-1\right)\left(x+1\right)}-\dfrac{x+2}{\left(x+1\right)^2}\right)\cdot\dfrac{\left(x-1\right)^2\cdot\left(x+1\right)^2}{4}\)
\(=\dfrac{x^2-x-2-x^2-x+2}{\left(x-1\right)\left(x+1\right)^2}\cdot\dfrac{\left(x-1\right)^2\cdot\left(x+1\right)^2}{4}\)
\(=\dfrac{-2x}{1}\cdot\dfrac{x-1}{4}=-\dfrac{x\left(x-1\right)}{2}\)
b: Để \(\dfrac{P-4}{5}=x\) thì P-4=5x
=>P=5x+4
\(\Leftrightarrow-\dfrac{x\left(x-1\right)}{2}=5x+4\)
=>-x2+x=10x+8
=>x2-x=-10x-8
=>x2+9x+8=0
=>x=-8(nhận) hoặc x=-1(loại)
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tim ginn cua bieu thuc P=\(\sqrt{\dfrac{x^2}{4}+\sqrt{ }x^2-4}+\sqrt{\dfrac{x^2}{4}-\sqrt{x^2-4}}\)
Cho cac bieu thuc :
\(A=\dfrac{\sqrt{x}+4}{\sqrt{x}+2},B=\left(\dfrac{\sqrt{x}}{\sqrt{x}+4}+\dfrac{4}{\sqrt{x}-4}\right):\dfrac{x+16}{\sqrt{x}+2}\)
a) Rut gon B ?
b) Tim cac gia tri nguyen cua x de cac gia tri cua bieu thuc B(A-1) la so nguyen.
Lần sau ghi dấu ra xíu nhé :v
a) Đặt \(\sqrt{x}=a\Rightarrow B=\left(\dfrac{a}{a+4}+\dfrac{4}{a-4}\right):\dfrac{a^2+16}{a+2}\)
Quy đồng,rút gọn : \(B=\dfrac{a+2}{a^2-16}\Rightarrow B=\dfrac{\sqrt{x}+2}{x-16}\)
b) \(B\left(A-1\right)=\dfrac{\sqrt{x}+2}{x-16}\left(\dfrac{\sqrt{x}+4}{\sqrt{x}+2}-1\right)=\dfrac{2}{x-16}\)
x - 16 là ước của 2 => \(x\in\left\{14;15;17;18\right\}\)
mới làm quen toán 9 ;v có gì k rõ ae chỉ bảo nhé :))
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1. Tim GTNN cua bieu thuc:
\(A=\dfrac{1}{2}\sqrt{x^2}+\sqrt{x^2-2x+1}\)
2. Rut gon bieu thuc:
a) \(A=\sqrt{29-4\sqrt{7}}+\sqrt{23+8\sqrt{7}}\)
b)\(B=\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}},voix>=2\)
Bài 2:
a: \(A=2\sqrt{7}-1+\left(\sqrt{7}+4\right)\)
\(=2\sqrt{7}-1+\sqrt{7}+4=3\sqrt{7}+3\)
b: \(B=\sqrt{x-1+2\sqrt{x-1}+1}+\sqrt{x-1-2\sqrt{x-1}+1}\)
\(=\sqrt{x-1}+1+1-\sqrt{x-1}=2\)
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