cho bieu thuc P=\(\dfrac{^{\left(a+3\right)^2}}{2a^26a}.\left(1-\dfrac{6a-18}{a^2-9}\right)\)
a.tim dieu kien xax dinh cua P
b.rut gon bieu thuc P
c.voi gia tri nao cua a thi P=0;P=1
Những câu hỏi liên quan
Cho bieu thuc:P=\(\dfrac{\left(a+3\right)^2}{a^2+3a}\times\left(1-\dfrac{6a-18}{a^2-9}\right)\)voi a ≠0;a≠ +-3
a)rut gon bieu thuc P
b)tim gia tri cua a de P= -2
c)tim cac gia tri nguyen cua a de bieu thuc P co gia tri nguyen
mng giup minh voi mai thi rui!
a: \(P=\dfrac{a+3}{a}\cdot\dfrac{a^2-9-6a+18}{\left(a-3\right)\left(a+3\right)}\)
\(=\dfrac{\left(a-3\right)^2}{a\left(a-3\right)}=\dfrac{a-3}{a}\)
b: Để P=-2 thì -2a=a-3
=>-3a=-3
=>a=1
c: Để P nguyên thì a-3 chia hết cho a
=>-3 chia hết cho a
mà a<>0; a<>3; a<>-3
nên \(a\in\left\{1;-1\right\}\)
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Cho bieu thuc A=\(\left(\dfrac{4}{x-\sqrt{x}}+\dfrac{\sqrt{x}}{\sqrt{x}-1}\right)\div\dfrac{1}{\sqrt{x}-1}\)
a/ Tim dieu kien cua x de bieu thuc A co gia tri xac dinh
b/ Rut gon A
c/ Tinh gia tri cua A khi x = \(4-2\sqrt{3}\)
d/ Tim gia tri nho nhat cua A
a. ĐKXĐ : x>1.
b. \(A=\left(\dfrac{4}{x-\sqrt{x}}+\dfrac{\sqrt{x}}{\sqrt{x}-1}\right):\dfrac{1}{\sqrt{x}-1}=\left[\dfrac{4}{\sqrt{x}\left(\sqrt{x}-1\right)}+\dfrac{\sqrt{x}}{\sqrt{x}-1}\right].\left(\sqrt{x}-1\right)=\dfrac{4+\sqrt{x}.\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}.\left(\sqrt{x}-1\right)=\dfrac{4+x}{\sqrt{x}}\)
c. Thay \(x=4-2\sqrt{3}\) vào A, ta có:
\(A=\dfrac{4+4-2\sqrt{3}}{\sqrt{4-2\sqrt{3}}}=\dfrac{8-2\sqrt{3}}{\sqrt{\left(\sqrt{3}-1\right)^2}}=\dfrac{8-2\sqrt{3}}{\sqrt{3}-1}=\dfrac{\left(8-2\sqrt{3}\right)\left(\sqrt{3}+1\right)}{3-1}=\dfrac{8\sqrt{3}+8-6-2\sqrt{3}}{2}=\dfrac{2+6\sqrt{3}}{2}=\dfrac{2\left(1+3\sqrt{3}\right)}{2}=1+3\sqrt{3}\)
Vậy giá trị của A tại \(x=4-2\sqrt{3}\) là \(1+3\sqrt{3}\).
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Cho phan thuc B=(3\y+3)+(1\y-3)-(18\9-y2)
a)Tim dieu kien cua y de gia tri cua bieu thuc B duoc xac dinh
b)Rut gon bieu thuc B
c)Tinh gia tri cua B de B co gia tri nguyen
Cho biểu thức A = x^2+3/x-2
a tim dieu kien cua x de gia tri cua bieu thuc khong xac dinh duoc
b. Voi nhung gia tri nao cua x thi bieu thuc A nhan gia tri la so a
cho bieu thuc A=\(\left(\frac{x-\sqrt{x}}{\sqrt{x}+1}+1\right):\left(\frac{x+\sqrt{x}}{\sqrt{x}+1}\right)\)
(x≥0;x≠1)
a.Tim x de bieu thuc A co nghia ? rut gon A ?
b. Tinh gia tri cua bieu thuc A tai x=7=4√3
cho 3 so a,b,c thoa man dieu kien : \(\left\{{}\begin{matrix}a+b+c=1\\\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\end{matrix}\right.\)
tinh gia tri cua bieu thuc T=\(a^2+b^2+c^2\)
\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\Rightarrow\dfrac{ab+bc+ca}{abc}=0\Rightarrow ab+bc+ca=0\)
T = a2 + b2 + c2 = (a + b+ c)2 - 2(ab + bc + ca) = 1 - 0 = 1
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cho 2 bieu thuc:
A=(\(\sqrt{20}\) -\(\sqrt{45}\) +3\(\sqrt{5}\) ).\(\sqrt{5}\) va B=\(\dfrac{x+1-2\sqrt{x}}{\sqrt{x}-1}\) +\(\dfrac{x+\sqrt{x}}{\sqrt{x}+1}\) (Dieu kien: x>0, x khac 1
a) Rut gon bieu thuc A va B
b)Tim cac gia tri cua x de gia tri cua bieu thuc A bang 2lan gia tri B
a: \(A=\left(2\sqrt{5}-3\sqrt{5}+3\sqrt{5}\right)\cdot\sqrt{5}=2\sqrt{5}\cdot\sqrt{5}=10\)
\(B=\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}-1}+\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\)
\(=\sqrt{x}-1+\sqrt{x}=2\sqrt{x}-1\)
b: A=2B
=>\(10=4\sqrt{x}-2\)
=>\(4\sqrt{x}=12\)
=>x=9(nhận)
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cho bieu thuc A=[x+2/x^2-x+x-2/x^2+x].x^2-1/x^2+2
a) tim dieu kien cua x de gia tri cua bieu thuc A duoc xac dinh
b) tinh gia tri cua bieu thuc A voi x = -200
a) \(A=\left[\dfrac{x+2}{x^2-x}+\dfrac{x-2}{x^2+x}\right].\dfrac{x^2-1}{x^2-x}\)
\(A=\left[\dfrac{x+2}{x\left(x-1\right)}+\dfrac{x-2}{x\left(x+1\right)}\right].\dfrac{x^2-1}{x^2+2}\)
\(A=\left[\dfrac{\left(x+2\right)\left(x+1\right)+\left(x-2\right)\left(x-1\right)}{x\left(x-1\right)\left(x+1\right)}\right].\dfrac{x^2-1}{x^2+2}\)
\(A=\left[\dfrac{x^2+2x+x+2+x^2-2x-x+2}{x\left(x-1\right)\left(x+1\right)}\right].\dfrac{x^2-1}{x^2+2}\)
\(A=\dfrac{2x^2+4}{x\left(x^2-1\right)}.\dfrac{x^2-1}{x^2+2}\)
\(A=\dfrac{2\left(x^2+2\right)\left(x^2-1\right)}{x\left(x^2-1\right)\left(x^2+2\right)}=\dfrac{2}{x}\)
b) Thay \(x=-200\) vào biểu thức \(A=\dfrac{2}{x}\) ta được :
\(A=\dfrac{2}{x}=\dfrac{2}{-200}=\dfrac{-2}{200}=\dfrac{-1}{100}\)
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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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