Tinh gia tri bieu thuc N=\(\frac{3a^2+6b^2-5c^2}{2a^2-4b^2+3c^2}\)biet 6a=4b=3c
Tính giá trị của biểu thức N=\(\dfrac{3a^2+6b^2-5c^2}{2a^2-4b^2+3c^2}\) biết 6a=4b=3c
Ta có:
6a = 4b = 3c
=> \(\dfrac{6a}{12}=\dfrac{4b}{12}=\dfrac{3c}{12}\)
=> \(\dfrac{a}{2}=\dfrac{b}{3}=\dfrac{c}{4}\)
=> \(\dfrac{a^2}{4}=\dfrac{b^2}{9}=\dfrac{c^2}{16}\)
Đặt \(\dfrac{a^2}{4}=\dfrac{b^2}{9}=\dfrac{c^2}{16}\)= k
=>\(\left\{{}\begin{matrix}a^2=4k\\b^2=9k\\c^2=16k\end{matrix}\right.\)
Thay \(\left\{{}\begin{matrix}a^2=4k\\b^2=9k\\c^2=16k\end{matrix}\right.\)vào biểu thức N ta được:
N = \(\dfrac{3a^2+6b^2-5c^2}{2a^2-4b^2+3c^2}\)
N = \(\dfrac{3.4k+6.9k-5.16k}{2.4k-4.9k+3.16k}\)
N = \(\dfrac{12k+54k-80k}{8k-36k+48k}\)
N = \(\dfrac{-14k}{20k}\)
N = \(\dfrac{-7}{10}\)
\(6a=4b=3c\Leftrightarrow\frac{a}{2}=\frac{b}{3}=\frac{c}{4}\)
Đặt \(\frac{a}{2}=\frac{b}{3}=\frac{c}{4}=k\Rightarrow\hept{\begin{cases}a=2k\\b=3k\\c=4k\end{cases}}\)
xong bạn thay vô biểu thức N rút gọn là ra
tính giá trị biểu thức : N = \(\dfrac{3a^2+6b^2-5c^2}{2a^2-4b^2+3c^2}\) biết 6a=4b=3c
help me
\(N=\dfrac{3a^2+6b^2-5c^2}{2a^2-4b^2+3c^2}\) (1)
Ta có:
\(6a=4b=3c\Rightarrow\dfrac{6a}{12}=\dfrac{4b}{12}=\dfrac{3c}{12}\Rightarrow\dfrac{a}{2}=\dfrac{b}{3}=\dfrac{c}{4}\)
Đặt \(\dfrac{a}{2}=\dfrac{b}{3}=\dfrac{c}{4}=k\Rightarrow\left\{{}\begin{matrix}a=2k\\b=3k\\c=4k\end{matrix}\right.\) (2)
Thay (2) vào (1) ta có:
\(\dfrac{3.\left(2k\right)^2+6.\left(3k\right)^2-5.\left(4k\right)^2}{2.\left(2k\right)^2-4.\left(3k\right)^2+3.\left(4k\right)^2}=\dfrac{3.4.k^2+6.9.k^2-5.16.k^2}{2.4.k^2-4.9.k^2+3.16.k^2}\)
\(=\dfrac{12k^2+54k^2-80k^2}{8k^2-36k^2+48k^2}=\dfrac{k^2.\left(12+54-80\right)}{k^2.\left(8-36+48\right)}=\dfrac{-14}{20}=\dfrac{-7}{10}\)
Vậy giá trị của biểu thức N là \(\dfrac{-7}{10}\)
Chúc bạn học tốt!!!
Tinh
Xy^3+4xy^3-3xy^3
(-4/5ab^2c)×(-20a^4b^3c)
Bai 2.tinh gia tri cua bieu thuc a=14x^2+5xy-2010y^2 tai x=-1;y=-2
xy3+4xy3-3xy3
=5xy3-3xy3 = 2xy3
tươg tự
Bài 2 : Thay zô có j kó đâu ==
Rút gọn biểu thức: \(P=\dfrac{3a^2+6b^2-5c^2}{2a^2-4b^2+3c^2}\)
Rút gọn biểu thức:
B=(-5c+3a-4b)-(3a-4b+7c)-(-12b-6a+15c)+(-3c+21a-10b)
C=-(-32b-12c+5a)+(2c-4b-23a)-(17a-16c-31b)-(-6b+3c)
B=(-5c+3a-4b)-(3a-4b+7c)-(-12b-6a+15c)+(-3c+21a-10b)
=-5c+3a-4b-3a+4b-7c+12b+6a-15c
=6a +12b -27c
C=-(-32b-12c+5a)+(2c-4b-23a)-(17a-16c-31b)-(-6b+3c)
=32b+12c-5a+2c-4b-23a-17a+16c+31b+6b-3c
=-45a+65b+9c
bn chỉ cần thêm bước đặt nhân tử chung thui nhé!!
cho a^3-4a^2b=2b^3-5ab^2 gia tri bieu thuc P=5a^2-4b^2+2ab/6a^2+2b^2-3ab
a3-4a2b=2b3-5ab2
=>(a3-3a2b+3ab2-b3)-(a2b+b3-2ab2)=0
=>(a-b)3-b(a2-2ab+b2)=0
=>(a-b)2(a-2b)=0
=> a-2b=0 (vì a#b#0 bạn thiếu điều kiện nha)
=>a=2b. Thay a=2b vào bt P ta đc P=1
cho a/2=b/3=c/4. tinh gia tri bieu thuc a^2+b^2+2c^2/a^2-4b^2+c^2
Giải:
Đặt \(\frac{a}{2}=\frac{b}{3}=\frac{c}{4}=k\)
\(\Rightarrow a=2k,b=3k,c=4k\)
Ta có: \(\frac{a^2+b^2+2c^2}{a^2-4b^2+c^2}\)
\(=\frac{\left(2k\right)^2+\left(3k\right)^2+2\left(4k\right)^2}{\left(2k\right)^2-4\left(3k\right)^2+\left(4k\right)^2}\)
\(=\frac{2^2.k^2+3^2.k^2+2.4^2.k^2}{2^2.k^2-4.3^2.k^2+4^2.k^2}\)
\(=\frac{4.k^2+9.k^2+32.k^2}{4.k^2-36.k^2+16.k^2}\)
\(=\frac{k^2.\left(4+9+32\right)}{k^2.\left(4-36+16\right)}\)
\(=\frac{45}{-16}\)
\(A=\frac{a^2+b^2+2c^2}{a^2-4b^2+c^2}\)
Đặt \(\frac{a}{2}=\frac{b}{3}=\frac{c}{4}=k\Rightarrow a=2k;b=3k;c=4k\)
Suy ra \(A=\frac{\left(2k\right)^2+\left(3k\right)^2+2\left(4k\right)^2}{\left(2k\right)^2-4\left(3k\right)^2+\left(4k\right)^2}=\frac{4k^2+9k^2+2\cdot16k^2}{4k^2-4\cdot9k^2+16k^2}\)
\(=\frac{k^2\left(4+9+32\right)}{k^2\left(4-36+16\right)}=\frac{45}{-16}=-\frac{45}{16}\)
Cho x+y=2, tinh gia tri cua bieu thuc:
M=3(x^2+y^2)-(x^3+y^3)+1
Bai 2:Cho a+b=5,tinh gia tri bieu thuc:
M=3a^2-2a+3b^2-2b+6ab+100