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a)F(x)=8+(-5x+3x)+(6x2 -3x2)+3x3
=8-2x+3x2+3x3
G(x)=-6+(12x2-9x2)+3x3
=-6+3x2+3x3
b)P(x)=8-2x+3x2+3x3-6+3x2+3x3
=(8-6)-2x+(3x2+3x2)+(3x3+3x3)
=2-2x+6x2+6x3
d)_thay \(\dfrac{1}{3}\) vào biểu thức F(x) ta có:
8-2.\(\dfrac{1}{3}\)+\(3.\left(\dfrac{1}{3}\right)^2\)+3.\(\left(\dfrac{1}{3}\right)^3\)
8-\(\dfrac{2}{3}\)+3.\(\dfrac{1}{9}\)+3.\(\dfrac{1}{27}\)
8-\(\dfrac{2}{3}\)+\(\dfrac{3}{9}\)+\(\dfrac{3}{27}\)
8-\(\dfrac{10}{27}\)
\(\dfrac{206}{27}\)
biểu thức G(x) tương tự chỗ nào có x bạn thay thành \(-\dfrac{1}{3}\)và tính thôi
c)mình chịu
a: \(\dfrac{x+1}{5}+\dfrac{x+1}{6}=\dfrac{x+1}{7}+\dfrac{x+1}{8}\)
\(\Leftrightarrow\left(x+1\right)\left(\dfrac{1}{5}+\dfrac{1}{6}-\dfrac{1}{7}-\dfrac{1}{8}\right)=0\)
=>x+1=0
hay x=-1
b: \(\Leftrightarrow\left(\dfrac{x-1}{2009}-1\right)+\left(\dfrac{x-2}{2008}-1\right)=\left(\dfrac{x-3}{2007}-1\right)+\left(\dfrac{x-4}{2006}-1\right)\)
=>x-2010=0
hay x=2010
c: \(\Leftrightarrow\dfrac{1}{x+2}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+10}+\dfrac{1}{x+10}-\dfrac{1}{x+17}=\dfrac{x}{\left(x+2\right)\left(x+17\right)}\)
\(\Leftrightarrow\dfrac{x}{\left(x+2\right)\left(x+17\right)}=\dfrac{x+17-x-2}{\left(x+2\right)\left(x+17\right)}\)
=>x=15
a/ \(\left|3x-1\right|=\left|5-2x\right|\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-1=5-2x\\3x-1=-5+2x\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}3x+2x=5+1\\3x-2x=-5+1\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}5x=6\\x=-4\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{6}{5}\\x=-4\end{matrix}\right.\)
Vậy ......
b/ \(\left|x+2\right|-\left|x+7\right|=0\)
\(\Leftrightarrow\left|x+2\right|=\left|x+7\right|\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=x+7\\x+2=-x-7\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-x=7-2\\x+x=-7-2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}0x=5\left(loại\right)\\2x=-9\end{matrix}\right.\)
\(\Leftrightarrow x=-\dfrac{9}{2}\)
Vậy ...............
c/ \(\left|2x-1\right|+x=2\)
\(\Leftrightarrow\left|2x-1\right|=2-x\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=2-x\\2x-1=-2+x\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+x=2+1\\2x-x=-2-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}3x=3\\x=-3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-3\end{matrix}\right.\)
Vậy ..
GIẢI CHI TIẾT RA GIÚP MK VS NHÉ!!!!MK RẤT CẢM ƠN
a.
\(\left(x+\frac{1}{2}\right)\times\left(x-\frac{3}{4}\right)=0\)
TH1:
\(x+\frac{1}{2}=0\)
\(x=-\frac{1}{2}\)
TH2:
\(x-\frac{3}{4}=0\)
\(x=\frac{3}{4}\)
Vậy \(x=-\frac{1}{2}\) hoặc \(x=\frac{3}{4}\)
b.
\(\left(\frac{1}{2}x-3\right)\times\left(\frac{2}{3}x+\frac{1}{2}\right)=0\)
TH1:
\(\frac{1}{2}x-3=0\)
\(\frac{1}{2}x=3\)
\(x=3\div\frac{1}{2}\)
\(x=3\times2\)
\(x=6\)
TH2:
\(\frac{2}{3}x+\frac{1}{2}=0\)
\(\frac{2}{3}x=-\frac{1}{2}\)
\(x=-\frac{1}{2}\div\frac{2}{3}\)
\(x=-\frac{1}{2}\times\frac{3}{2}\)
\(x=-\frac{3}{4}\)
Vậy \(x=6\) hoặc \(x=-\frac{3}{4}\)
c.
\(\frac{2}{3}-\frac{1}{3}\times\left(x-\frac{3}{2}\right)-\frac{1}{2}\times\left(2x+1\right)=5\)
\(\frac{2}{3}-\frac{1}{3}x+\frac{1}{2}-x-\frac{1}{2}=5\)
\(\left(\frac{1}{2}-\frac{1}{2}\right)-\left(\frac{1}{3}x+x\right)=5-\frac{2}{3}\)
\(-\frac{4}{3}x=\frac{13}{3}\)
\(x=\frac{13}{3}\div\left(-\frac{4}{3}\right)\)
\(x=\frac{13}{3}\times\left(-\frac{3}{4}\right)\)
\(x=-\frac{13}{4}\)
d.
\(4x-\left(x+\frac{1}{2}\right)=2x-\left(\frac{1}{2}-5\right)\)
\(4x-x-\frac{1}{2}=2x-\frac{1}{2}+5\)
\(4x-x-2x=\frac{1}{2}-\frac{1}{2}+5\)
\(x=5\)
Đề sai bạn nhé. Đưa dữ kiện 3 ẩn bắt tính biểu thức chứa 2 ẩn làm sao làm được ?
Bạn kiểm tra lại nha
\(a.M+(5x^2-2xy)=6x^2+9xy-y^2
\)
\(M=(6x^2+9xy-y^2)-(5x^2-2xy)\)
\(M=6x^2+9xy-y^2-5x^2+2xy\)
\(M=(6x^2-5x^2)+(9xy+2xy)-y^2\)
\(M=x^2+11xy-y^2\)
Vậy \(M=x^2+11xy-y^2\)
\(b.M+(3x^2y-2xy^3)=2x^2y-4xy^3\)
\(M=(2x^2y-4xy^3)-(3x^2-2xy^3)\)
\(M=
\) \(2x^2-4xy^3-3x^2+2xy^3\)
\(M=(2x^2-3x^2)+(-4xy^3+2xy^3)\)
\(M=-x^2-2xy^3\)
Vậy \(M=-x^2-2xy^3\)
a) M + (5x\(^2\) - 2xy) = 6x\(^2\) + 9xy - y\(^2\)
=> M = (6x\(^2\) + 9xy - y\(^2\)) - (5x\(^2\) - 2xy)
M = 6x\(^2\) + 9xy - y\(^2\) - 5x\(^2\) + 2xy
M = (6x\(^2\) - 5x\(^2\)) + (9xy + 2xy) - y\(^2\)
M = 1x\(^2\) + 11xy - y\(^2\)
2.Áp dụng tc dãy tỉ số bằng nhau ta có:
\(\frac{a+b+c}{a+b-c}=\frac{a-b+c}{a-b-c}=\frac{a+b+c-a+b-c}{a+b-c-a+b+c}=\frac{2b}{2b}=1\)
\(\Rightarrow a+b+c=a+b-c\)
\(\Rightarrow a+b+c-a-b+c=0\)
\(\Rightarrow2c=0\)
\(\Rightarrow c=0\)
Vậy c=0
BT5: Ta có: f(1)=1.a+b=1 =>a+b=1 (1)
f(2)=2a+b=4 (2)
Trừ (1) cho (2) ta có: 2a+b-a-b=4-1 => a=3
Với a=3 thay vào (1) ta có: 3+b=1 => b=-2
Vậy a=3, b=-2
Cái này dễ lắm. Mình giải luôn nhé!
a) \(\left[{}\begin{matrix}\dfrac{1}{7}x-\dfrac{2}{7}=0\Leftrightarrow x=\dfrac{2}{7}:\dfrac{1}{7}\Leftrightarrow x=2\\-\dfrac{1}{5}x+\dfrac{3}{5}=0\Leftrightarrow x=-\dfrac{3}{5}:\left(-\dfrac{1}{5}\right)\Leftrightarrow x=3\\\dfrac{1}{3}x+\dfrac{4}{3}=0\Leftrightarrow x=-\dfrac{4}{3}:\dfrac{1}{3}\Leftrightarrow x=-4\end{matrix}\right.\)
Vậy x=2 hoặc x=3 hoặc x=-4
b)\(x\left(\dfrac{1}{6}+\dfrac{1}{10}-\dfrac{4}{15}\right)+1=0\)
\(x.0+1=0\)
\(1=0\) ( vô lí)
Vậy không có giá trị của x nào thỏa mãn
a,f(x)=0
<=>3x-6=0
<=>3x=6
<=>x=2
Vậy nghiệm của f(x)=2
sử dụng TC phân phối Thu Trang