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\(\text{a) }3-2\left|4x-5\right|=\dfrac{2}{6}\\ \Leftrightarrow2\left|4x-5\right|=\dfrac{8}{3}\\ \Leftrightarrow\left|4x-5\right|=\dfrac{4}{3}\\ \Leftrightarrow4x-5=-\dfrac{4}{3}\text{ hoặc :}\\ 4x-5=-\dfrac{4}{3}\\ \Leftrightarrow\left[{}\begin{matrix}4x-5=-\dfrac{4}{3}\\4x-5=\dfrac{4}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}4x=\dfrac{11}{3}\\4x=\dfrac{19}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{11}{12}\\x=\dfrac{19}{12}\end{matrix}\right.\\ \text{Vậy }x=\dfrac{11}{12}\text{ hoặc }x=\dfrac{19}{12}\)
b) \(\left|4-7x\right|-\dfrac{3}{2}:5=\left|-1\dfrac{1}{3}\right|\)
\(\left|4-7x\right|-\dfrac{3}{10}=\dfrac{4}{3}\)
\(\left|4-7x\right|=\dfrac{49}{30}\) (*)
+) Nếu 4 - 7x \(\ge\) 0 \(\Rightarrow x\le\dfrac{4}{7}\)
PT (*) \(\Leftrightarrow4-7x=\dfrac{49}{30}\)
\(-7x=-\dfrac{71}{30}\)
x = \(\dfrac{71}{210}\) (t/m)
+) Nếu \(4-7x< 0\Rightarrow x>\dfrac{4}{7}\)
Pt (*) \(\Leftrightarrow-4+7x=\dfrac{49}{30}\)
x = \(\dfrac{169}{210}\) (t/m)
Vậy x=\(\dfrac{71}{210}\) hoặc x = \(\dfrac{169}{210}\)
c, \(\left(7-3x\right)\left(2x+1\right)=0\)
=> \(7-3x=0\) hoặc \(2x+1=0\)
\(3x=7-0\) hoặc \(2x=0-1\)
\(3x=7\) hoặc \(2x=-1\)
\(x=7:3\) hoặc \(x=-1:2\)
\(x=\dfrac{7}{3}\) hoặc \(x=-0,5\)
Vậy, \(x\in\left\{\dfrac{7}{3};-0,5\right\}\)
\(d,x-5\sqrt{x}=0\)
\(ĐKXĐ:x\ge0\)
\(\Leftrightarrow\sqrt{x}\left(\sqrt{x}-5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x}=0\\\sqrt{x}-5=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\\sqrt{x}=5\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=25\end{cases}}\)(Thỏa mãn ĐKXĐ)
Vậy...
Bài 1:
|\(x\)| = 1 ⇒ \(x\) \(\in\) {-\(\dfrac{1}{3}\); \(\dfrac{1}{3}\)}
A(-1) = 2(-\(\dfrac{1}{3}\))2 - 3.(-\(\dfrac{1}{3}\)) + 5
A(-1) = \(\dfrac{2}{9}\) + 1 + 5
A (-1) = \(\dfrac{56}{9}\)
A(1) = 2.(\(\dfrac{1}{3}\) )2- \(\dfrac{1}{3}\).3 + 5
A(1) = \(\dfrac{2}{9}\) - 1 + 5
A(1) = \(\dfrac{38}{9}\)
|y| = 1 ⇒ y \(\in\) {-1; 1}
⇒ (\(x;y\)) = (-\(\dfrac{1}{3}\); -1); (-\(\dfrac{1}{3}\); 1); (\(\dfrac{1}{3};-1\)); (\(\dfrac{1}{3};1\))
B(-\(\dfrac{1}{3}\);-1) = 2.(-\(\dfrac{1}{3}\))2 - 3.(-\(\dfrac{1}{3}\)).(-1) + (-1)2
B(-\(\dfrac{1}{3}\); -1) = \(\dfrac{2}{9}\) - 1 + 1
B(-\(\dfrac{1}{3}\); -1) = \(\dfrac{2}{9}\)
B(-\(\dfrac{1}{3}\); 1) = 2.(-\(\dfrac{1}{3}\))2 - 3.(-\(\dfrac{1}{3}\)).1 + 12
B(-\(\dfrac{1}{3};1\)) = \(\dfrac{2}{9}\) + 1 + 1
B(-\(\dfrac{1}{3}\); 1) = \(\dfrac{20}{9}\)
B(\(\dfrac{1}{3};-1\)) = 2.(\(\dfrac{1}{3}\))2 - 3.(\(\dfrac{1}{3}\)).(-1) + (-1)2
B(\(\dfrac{1}{3}\); -1) = \(\dfrac{2}{9}\) + 1 + 1
B(\(\dfrac{1}{3}\); -1) = \(\dfrac{20}{9}\)
B(\(\dfrac{1}{3}\); 1) = 2.(\(\dfrac{1}{3}\))2 - 3.(\(\dfrac{1}{3}\)).1 + (1)2
B(\(\dfrac{1}{3}\); 1) = \(\dfrac{2}{9}\) - 1 + 1
B(\(\dfrac{1}{3}\);1) = \(\dfrac{2}{9}\)
bài 1:
|x| = \(\dfrac{1}{3}\) => x = \(\pm\)\(\dfrac{1}{3}\) |y| = 1 => y = \(\pm\)1
a
+) A = 2x\(^2\) - 3x + 5
= 2\(\left(\dfrac{1}{3}\right)^2\) - 3.\(\dfrac{1}{3}\) +5 = 2.\(\dfrac{1}{9}\) - 1 + 5
= \(\dfrac{2}{9}\) - 1 + 5 = \(\dfrac{2-9+45}{9}\) = \(\dfrac{38}{9}\)
+) A = 2x\(^2\) - 3x + 5
= 2\(\left(\dfrac{-1}{3}\right)^2\) - 3\(\left(\dfrac{-1}{3}\right)\) + 5
= 2.\(\dfrac{1}{9}\) - (-1) + 5 = \(\dfrac{2}{9}\) + 1 +5
= \(\dfrac{2+9+45}{9}\) = \(\dfrac{56}{9}\)
b) +) B = 2x\(^2\) - 3xy + y\(^2\)
= 2\(\left(\dfrac{1}{3}\right)^2\) - 3.\(\dfrac{1}{3}\).1 + 1\(^2\)
= 2.\(\dfrac{1}{9}\) - 1 + 1 = \(\dfrac{2}{9}\) - 1 + 1
= \(\dfrac{2-9+9}{9}\) = \(\dfrac{2}{9}\)
+) B = 2x\(^2\) - 3xy + y\(^2\)
= 2\(\left(\dfrac{-1}{3}\right)\)\(^2\) - 3\(\left(\dfrac{-1}{3}\right)\). 1 + 1\(^2\)
= 2.\(\dfrac{1}{9}\) - (-1) + 1 = \(\dfrac{2}{9}\) + 1 + 1
= \(\dfrac{2+9+9}{9}\) = \(\dfrac{20}{9}\)
bài 3
x.y.z = 2 và x + y + z = 0
A = ( x + y )( y +z )( z + x )
= x + y . y + z . z + x = ( x + y + z ) + ( x . y . z )
= 0 + 2 = 2
bài 4
a) | 2x - \(\dfrac{1}{3}\) | - \(\dfrac{1}{3}\) = 0 => | 2x - \(\dfrac{1}{3}\) | = \(\dfrac{1}{3}\)
=> 2x - \(\dfrac{1}{3}\) = \(\pm\) \(\dfrac{1}{3}\)
+) 2x - \(\dfrac{1}{3}\)= \(\dfrac{1}{3}\)
=> 2x = \(\dfrac{1}{3}\) + \(\dfrac{1}{3}\) = \(\dfrac{2}{3}\)
x = \(\dfrac{2}{3}\) : 2 = \(\dfrac{2}{3}\) . \(\dfrac{1}{2}\) = \(\dfrac{1}{3}\)
+) 2x - \(\dfrac{1}{3}\) = \(\dfrac{-1}{3}\)
2x = \(\dfrac{-1}{3}\) + \(\dfrac{1}{3}\) = 0
x = 0 : 2 = 2
Dạng 1:
a) $4x+9=4x+\frac{9}{4}.4=4(x+\frac{9}{4}\Rightarrow$ Nghiệm là $-\frac{9}{4}$
b) $-5x+6=-5x+(-5).(-\frac{6}{5})=-5(x-\frac{6}{5})\Rightarrow$ Nghiệm là $\frac{6}{5}$
c) $7-2x=-2x+7=-2x+(-2).(-\frac{7}{2})=-2(x-\frac{7}{2})\Rightarrow$ Nghiệm là $\frac{7}{2}$
d) $2x+5=2x+2.\frac{5}{2}=2.(x+\frac{5}{2})\Rightarrow$ Nghiệm là $-\frac{5}{2}$
e) $2x+6=2x+2.3=2(x+3)\Rightarrow$ Nghiệm là -3
g) $3x-\frac{1}{4}=3x-3.(\frac{1}{12})=3(x-\frac{1}{12})\Rightarrow$ Nghiệm là $\frac{1}{12}$
h) $3x-9=3x-3.3=3(x-3)\Rightarrow$ Nghiệm là 3
k) $-3x-\frac{1}{2}=-3x-3.(\frac{1}{6})=-3(x+\frac{1}{6})\Rightarrow$ Nghiệm là $-\frac{1}{6}$
m) $-17x-34=-17x-17.2=-17(x+2)\Rightarrow$ Nghiệm là -2
n) $2x-1=2x+2.(-\frac{1}{2})=3(x-\frac{1}{2})\Rightarrow$ Nghiệm là $\frac{1}{2}$
q) $5-3x=-3x+5=-3x+(-3).(-\frac{5}{3})=-3(x-\frac{5}{3})\Rightarrow$ Nghiệm là $\frac{5}{3}$
p) $3x-6=3x+3.(-2)=3(x-2)\Rightarrow$ Nghiệm là 2
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