Bài này là một bài cơ bản, bạn nên tự làm

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/ \(\left|3x-1\right|=\left|5-2x\right|\)

\(\Leftrightarrow\left[{}\begin{matrix}3x-1=5-2x\\3x-1=-5+2x\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}3x+2x=5+1\\3x-2x=-5+1\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}5x=6\\x=-4\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{6}{5}\\x=-4\end{matrix}\right.\)

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 ..

Tự lực suy nghĩ mà làm một lần đi, đừng hỏi nữa.

Mình có hỏi nữa đâu!

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

\(a.M+(5x^2-2xy)=6x^2+9xy-y^2 \)
\(M=(6x^2+9xy-y^2)-(5x^2-2xy)\)
\(M=6x^2+9xy-y^2-5x^2+2xy\)
\(M=(6x^2-5x^2)+(9xy+2xy)-y^2\)
\(M=x^2+11xy-y^2\)
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\(^2\)

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.

\(4x-\left(x+\frac{1}{2}\right)=2x-\left(\frac{1}{2}-5\right)\)

\(4x-x-\frac{1}{2}=2x-\frac{1}{2}+5\)

\(4x-x-2x=\frac{1}{2}-\frac{1}{2}+5\)

\(x=5\)

a/ \(\left|-x\right|=1,5\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1,5\\x=-1,5\end{matrix}\right.\)

Vậy .....

b/ \(\left|x+\dfrac{1}{2}\right|=2\dfrac{1}{2}\)

\(\Leftrightarrow\left|x+\dfrac{1}{2}\right|=\dfrac{5}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{5}{2}\\x+\dfrac{1}{2}=-\dfrac{5}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)

Vậy ....

c/ \(\left|0,5-x\right|=\left|-0,5\right|\)

\(\left|0,5-x\right|=0,5\)

\(\Leftrightarrow\left[{}\begin{matrix}0,5-x=0,5\\0,5-x=-0,5\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)

Vậy ...

Cảm ơn bn nha.

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

Tìm x dễ thì tự làm nha:

\(\dfrac{x+4}{2000}+\dfrac{x+3}{2001}=\dfrac{x+2}{2002}+\dfrac{x+1}{2003}\)

\(\Rightarrow\dfrac{x+4}{2000}+\dfrac{x+3}{2001}-\dfrac{x+2}{2002}-\dfrac{x+1}{2003}=0\)

\(\Rightarrow\left(\dfrac{x+4}{2000}+1\right)+\left(\dfrac{x+3}{2001}+1\right)-\left(\dfrac{x+2}{2002}+1\right)-\left(\dfrac{x+1}{2003}\right)=0\)\(\Rightarrow\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}-\dfrac{x+2004}{2002}-\dfrac{x+2004}{2003}=0\)

\(\Rightarrow\left(x+2004\right)\left(\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\right)=0\)

\(\Rightarrow x+2004=0\Rightarrow x=-2004\)