tinh gia tri cua bieu thuc :
\(\dfrac{x^3}{8}+\dfrac{x^2y}{4}+\dfrac{xy^2}{6}+\dfrac{y^3}{27}taix=-8;y=6\)
Những câu hỏi liên quan
cho x,y,z la cac so nguyen duong thoa man \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=2015\)
tinh gia tri lon nhat cua bieu thuc P=\(\dfrac{xy}{x^3+y^3}+\dfrac{yz}{y^3+z^3}+\dfrac{zx}{z^{3+x^3}}\)
gia tri cua bieu thuc : B=\(\dfrac{-1:1\dfrac{1}{15}}{3\dfrac{1}{8}:6\dfrac{2}{3}}\): \(\dfrac{4\dfrac{7}{8}:13}{5:1\dfrac{7}{8}}\): \(\dfrac{1}{3^2}\)
1.Tinh gia tri cua bieu thuc
\(\dfrac{1}{2}+\dfrac{1}{3}\).x-\(\dfrac{1}{6}\).x voi x=\(\dfrac{-3}{5}\)
Thay \(x=-\dfrac{3}{5}\) vào biểu thức ta có:
\(\dfrac{1}{2}+\dfrac{1}{3}.\dfrac{-3}{5}-\dfrac{1}{6}.\dfrac{-3}{5}\)
\(=\dfrac{1}{2}+\dfrac{-3}{5}.\left(\dfrac{1}{3}-\dfrac{1}{6}\right)=\dfrac{1}{2}+\dfrac{-3}{5}.\dfrac{1}{6}\)
\(=\dfrac{1}{2}+\dfrac{-1}{10}=\dfrac{2}{5}\)
Chúc bạn học tốt!!!
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đóng góp một cách khác:
đặt biểu thức trên là A.
\(A=\dfrac{1}{2}+\dfrac{1}{3}x-\dfrac{1}{6}x=\dfrac{1}{2}+\dfrac{x}{6}\)
Thay \(x=-\dfrac{3}{5}\) vào biểu thức A, ta có:
\(A=\dfrac{1}{2}+\dfrac{-\dfrac{3}{5}}{6}\\ =\dfrac{1}{2}-\dfrac{1}{10}\\ =\dfrac{5}{10}-\dfrac{1}{10}\\ =\dfrac{4}{10}\\ =\dfrac{2}{5}\)
Vậy giá trị biểu thức A tại \(x=-\dfrac{3}{5}\) là \(\dfrac{2}{5}\)
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Thay \(x=-\dfrac{3}{5}\) vào biểu thức ta có:
\(\dfrac{1}{2}+\dfrac{1}{3}.\dfrac{-3}{5}-\dfrac{1}{6}.\dfrac{-3}{5}\)
\(=\dfrac{1}{2}+\dfrac{-3}{5}.\left(\dfrac{1}{3}-\dfrac{1}{6}\right)=\dfrac{1}{2}+\dfrac{-3}{5}.\dfrac{1}{6}\)
\(=\dfrac{1}{2}+\dfrac{-1}{10}=\dfrac{2}{5}\)
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Xem thêm câu trả lời
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 bieu thuc \(P=\left(\dfrac{3}{x-1}-\dfrac{1}{\sqrt{x}+1}\right):\dfrac{1}{\sqrt{x}+1}\)
a.Neu dkxd va rut gon bieu thuc P
b.Tim cac gia tri cua x de \(P=\dfrac{5}{4}\)
c.Tim gia tri nho nhat cua bieu thuc :\(M=\dfrac{x+12}{\sqrt{x}-1}\cdot\dfrac{1}{P}\)
a)ĐKXĐ:x>0
P=\(\left(\frac{3}{x-1}-\frac{1}{\sqrt{x}+1}\right):\frac{1}{\sqrt{x}+1}\left(vớix>0\right)\)
=\(\left[\frac{3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{1}{\sqrt{x}+1}\right]:\frac{1}{\sqrt{x}+1}\)
=\(\left[\frac{3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right]:\frac{1}{\sqrt{x}+1}\)
= \(\left[\frac{3-\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right]:\frac{1}{\sqrt{x}+1}\)
=\(\frac{4-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\frac{\sqrt{x}+1}{1}\)
=\(\frac{4-\sqrt{x}}{\sqrt{x}-1}\)
b)Để P=\(\frac{5}{4}\left(vớix>0\right)\)
\(\Leftrightarrow\frac{4-\sqrt{x}}{\sqrt{x}-1}=\frac{5}{4}\)
\(\Leftrightarrow\frac{4-\sqrt{x}}{\sqrt{x}-1}-\frac{5}{4}=0\)
\(\Leftrightarrow\frac{4\left(4-\sqrt{x}\right)}{4\left(\sqrt{x}-1\right)}-\frac{5\left(\sqrt{x}-1\right)}{4\left(\sqrt{x}-1\right)}=0\)
\(\Rightarrow16-4\sqrt{x}-5\sqrt{x}+5=0\)
\(\Leftrightarrow21-9\sqrt{x}=0\)
\(\Leftrightarrow-9\sqrt{x}=-21\)
\(\Leftrightarrow\sqrt{x}=\frac{7}{3}\)
\(\Leftrightarrow x=\frac{21}{9}\)
Vậy:Để P=\(\frac{5}{4}\)thì x=\(\frac{21}{9}\)
c)Còn phần c thì mik chịu
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 A = \(\left(\dfrac{\sqrt{x}+\sqrt{y}}{1-\sqrt{xy}}+\dfrac{\sqrt{x}-\sqrt{y}}{1+\sqrt{xy}}\right):\left(1+\dfrac{x+y+2xy}{1-xy}\right)\)
a, Rut gon bieu thuc A
b, Tinh gia tri cua A khi x = \(\dfrac{1}{1+\sqrt{2}}\)
c, Tim Max A
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a: \(A=\dfrac{\sqrt{x}+x\sqrt{y}+\sqrt{y}+y\sqrt{x}+\sqrt{x}-x\sqrt{y}-\sqrt{y}+y\sqrt{x}}{1-xy}:\dfrac{1-xy+x+y+2xy}{1-xy}\)
\(=\dfrac{2\sqrt{x}+2y\sqrt{x}}{x+y+xy+1}\)
\(=\dfrac{2\sqrt{x}\left(y+1\right)}{\left(x+1\right)\left(y+1\right)}=\dfrac{2\sqrt{x}}{x+1}\)
b: \(x=\dfrac{1}{\sqrt{2}+1}=\sqrt{2}-1\)
\(A=\dfrac{2\sqrt{\sqrt{2}-1}}{\sqrt{2}-1+1}=\sqrt{2\left(\sqrt{2}-1\right)}\)
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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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