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tớ rút kinh nghiệm từ 1 bn nha !
\(a,\left|x+5\right|=\left|3x-2\right|\)
\(\Rightarrow\orbr{\begin{cases}x+5=3x-2\\x+5=-3x-2\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x+5-3x+2=0\\x+5+3x+2=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}-2x+7=0\\4x+7=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}-2x=-7\\4x=-7\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{7}{2}\\x=-\frac{7}{4}\end{cases}}\)
còn đâu bn tự lm nha !!! , bn lm mới có ý nghĩa , cố lên !!! , sai đâu nhắc sửa nha !
a) |x+5|=|3x-2|
\(\Rightarrow\orbr{\begin{cases}x+5=3x-2\\x+5=-3x+2\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}5+2=3x-x\Rightarrow7=2x\Rightarrow x=\frac{7}{2}\\x+3x=2-5\Rightarrow4x=-3\Rightarrow x=\frac{-3}{4}\end{cases}}\)
Vay..................................................................................

a) \(A=2x^2+3x+1=\left(2x^2+2x\right)+\left(x+1\right)\)
\(=2x\left(x+1\right)+\left(x+1\right)=\left(x+1\right)\left(2x+1\right)\)
Ta có: \(\left|x\right|=\frac{1}{2}\)\(\Leftrightarrow\orbr{\begin{cases}x=\frac{-1}{2}\\x=\frac{1}{2}\end{cases}}\)
TH1: Nếu \(x=\frac{-1}{2}\)\(\Rightarrow A=\left(\frac{-1}{2}+1\right)\left(2.\frac{-1}{2}+1\right)=\left(\frac{-1}{2}+1\right)\left(-1+1\right)=0\)
TH2: Nếu \(x=\frac{1}{2}\)\(\Rightarrow A=\left(\frac{1}{2}+1\right)\left(2.\frac{1}{2}+1\right)=\frac{3}{2}.\left(1+1\right)=\frac{3}{2}.2=3\)
Vậy \(A=0\)hoặc \(A=3\)
b) Thay \(x=-1\)và \(y=2\)vào biểu thức ta được:
\(B=\left(-1\right)^2.2-3.\left(-1\right).2^2+\left(-1\right)^2.2^2=2+12+4=18\)

a)
\(\sqrt{x}=7\Rightarrow x=49\)
b) \(\sqrt{2}-3x=4\Rightarrow3x=\sqrt{2}-4\)
\(x=\frac{\sqrt{2-4}}{3}\)
c)suy ra \(\frac{x+1}{2}=\frac{3}{2}\)suy ra x+1=3 suy ra x=2

a) Vì x< 0 nên x= \(-\sqrt{7}\)
b) x-2 =\(\sqrt{2}\)hoặc x-2 = -\(\sqrt{2}\)
suy ra x= \(\sqrt{2}\)+2 hoặc x= \(-\sqrt{2}\)+2
c)
x+\(\sqrt{3}\) =\(\sqrt{5}\)hoặc x+\(\sqrt{3}\) = -\(\sqrt{5}\)
suy ra x= \(\sqrt{5}-\sqrt{3}\)hoặc x= \(-\sqrt{5}-\sqrt{3}\)
Các bạn tự kết luận nhé

1) 1/x-1/y
=y/xy-x/xy
=y-x/xy
= - (x-y)/xy
= -1 (vì x-y=xy)
2)
(x- 1/2)*(y+1/3)*(z-2)=0
=> x-1/2 = 0 hoac y+1/3=0 hoac z-2=0
th1 :x-1/2=0 => x=1/2
x+2=y+3=z+4
mà x=1/2 => y= -1/2 ; z=-3/2
th2: y+1/3=0
th3 : z-2=0
(tự làm nha)
1) Với x,y khác 0, Ta có
\(\frac{1}{x}-\frac{1}{y}=\frac{y-x}{xy}=-\left(\frac{x-y}{xy}\right)=-\left(\frac{xy}{xy}\right)=-1\)
Vậy \(\frac{1}{x}-\frac{1}{y}=-1\)
2) Ta có:
\(\left(x-\frac{1}{2}\right)\left(y+\frac{1}{3}\right)\left(z-2\right)=0\)
Trường hợp 1: x - 1/2 = 0 => x = 1/2 \(\Rightarrow\hept{\begin{cases}y=\frac{1}{2}+2-3=-\frac{1}{2}\\z=\frac{1}{2}+2-4=-\frac{3}{2}\end{cases}}\)
Trường hợp 2: y + 1/3 = 0 => y = -1/3 \(\Rightarrow\hept{\begin{cases}x=-\frac{1}{3}+3-2=\frac{2}{3}\\z=-\frac{1}{3}+3-4=-\frac{4}{3}\end{cases}}\)
Trường hợp 3: z - 2 = 0 => z = 2 \(\Rightarrow\hept{\begin{cases}x=2+4-2=4\\y=2+4-3=3\end{cases}}\)
Vậy......

1) \(\left|x-\frac{3}{5}\right|< \frac{1}{3}\)
\(\Rightarrow\orbr{\begin{cases}x-\frac{3}{5}< \frac{1}{3}\\x-\frac{3}{5}< -\frac{1}{3}\end{cases}}\Rightarrow\orbr{\begin{cases}x< \frac{1}{3}+\frac{3}{5}\\x< \frac{-1}{3}+\frac{3}{5}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x< \frac{5}{15}+\frac{9}{15}\\x< \frac{-5}{15}+\frac{9}{15}\end{cases}}\Rightarrow\orbr{\begin{cases}x< \frac{14}{15}\\x< \frac{4}{15}\end{cases}}\)
vay \(\orbr{\begin{cases}x< \frac{14}{15}\\x< \frac{4}{15}\end{cases}}\)
2) \(\left|x+\frac{11}{2}\right|>\left|-5,5\right|\)
\(\left|x+\frac{11}{2}\right|>5,5\)
\(\Rightarrow\orbr{\begin{cases}x+\frac{11}{2}>\frac{11}{2}\\x+\frac{11}{2}>-\frac{11}{2}\end{cases}}\Rightarrow\orbr{\begin{cases}x>\frac{11}{2}-\frac{11}{2}\\x>\frac{-11}{2}-\frac{11}{2}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x>0\\x>-11\end{cases}}\)
vay \(\orbr{\begin{cases}x>0\\x>-11\end{cases}}\)
3) \(\frac{2}{5}< \left|x-\frac{7}{5}\right|< \frac{3}{5}\)
\(\Rightarrow\left|x-\frac{7}{5}\right|>\frac{2}{5}\) va \(\left|x-\frac{7}{5}\right|< \frac{3}{5}\)
\(\Rightarrow\orbr{\begin{cases}x-\frac{7}{5}>\frac{2}{5}\\x-\frac{7}{5}>\frac{-2}{5}\end{cases}}\Rightarrow\orbr{\begin{cases}x>\frac{2}{5}+\frac{7}{5}\\x>\frac{-2}{5}+\frac{7}{5}\end{cases}}\)va \(\orbr{\begin{cases}x-\frac{7}{5}< \frac{3}{5}\\x-\frac{7}{5}< \frac{-3}{5}\end{cases}}\Rightarrow\orbr{\begin{cases}x< \frac{3}{5}+\frac{7}{5}\\x< \frac{-3}{5}+\frac{7}{5}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x>\frac{9}{5}\\x>1\end{cases}}\)va \(\orbr{\begin{cases}x< 2\\x< \frac{4}{5}\end{cases}}\)
vay ....

a, A lớn nhất khi 7x la nguyên dương nho nhất
\(\Rightarrow7x=1\)
\(\Rightarrow x=\frac{1}{7}\)
\(b,B=\frac{10+4-x}{4-x}\)
\(B=\frac{10}{4-x}+1\)
b lon nhat khi 4-xla nguyen duong nho nhat
\(\Rightarrow4-x=1\)
\(\Rightarrow x=4-1=3\)
\(c,C=\frac{27-2x}{12-x}=\frac{3+24-2x}{12-x}=\frac{3}{12-x}+2\)
c lon nhat khi 12-x la nguyen duong nho nhat
\(\Rightarrow12-x=1\Rightarrow x=11\)
|x - 1| + 2|x| = |-11|
|x - 1| + 2|x| = 11
TH1: \(x\ge0\)\(\Rightarrow\left|x-1\right|=x-1\)và \(\left|x\right|=x\)
vậy x - 1 + 2x = 11
3x = 12
x = 4
TH2: \(x< 0\)\(\Rightarrow\left|x-1\right|=1-x\)và \(\left|x\right|=-x\)
vậy 1 - x + 2(-x) = 11
1 - x - 2x = 11
-3x = 10
x = -10/3