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Bài 1:
a) Ta có: \(\dfrac{17}{6}-x\left(x-\dfrac{7}{6}\right)=\dfrac{7}{4}\)
\(\Leftrightarrow\dfrac{17}{6}-x^2+\dfrac{7}{6}x-\dfrac{7}{4}=0\)
\(\Leftrightarrow-x^2+\dfrac{7}{6}x+\dfrac{13}{12}=0\)
\(\Leftrightarrow-12x^2+14x+13=0\)
\(\Delta=14^2-4\cdot\left(-12\right)\cdot13=196+624=820\)
Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{14-2\sqrt{205}}{-24}=\dfrac{-7+\sqrt{205}}{12}\\x_2=\dfrac{14+2\sqrt{2015}}{-24}=\dfrac{-7-\sqrt{205}}{12}\end{matrix}\right.\)
b) Ta có: \(\dfrac{3}{35}-\left(\dfrac{3}{5}-x\right)=\dfrac{2}{7}\)
\(\Leftrightarrow\dfrac{3}{5}-x=\dfrac{3}{35}-\dfrac{10}{35}=\dfrac{-7}{35}=\dfrac{-1}{5}\)
hay \(x=\dfrac{3}{5}-\dfrac{-1}{5}=\dfrac{3}{5}+\dfrac{1}{5}=\dfrac{4}{5}\)
\(a,\Leftrightarrow\left|x+\dfrac{2}{5}\right|=\dfrac{7}{4}\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{2}{5}=\dfrac{7}{4}\left(x\ge-\dfrac{2}{5}\right)\\x+\dfrac{2}{5}=-\dfrac{7}{4}\left(x< -\dfrac{2}{5}\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{27}{20}\left(tm\right)\\x=-\dfrac{43}{20}\left(tm\right)\end{matrix}\right.\)
\(b,\Leftrightarrow\left|x-\dfrac{13}{10}\right|=\dfrac{13}{10}\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{13}{10}=\dfrac{13}{10}\left(x\ge\dfrac{13}{10}\right)\\x-\dfrac{13}{10}=-\dfrac{13}{10}\left(x< \dfrac{13}{10}\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{13}{5}\left(tm\right)\\x=0\left(tm\right)\end{matrix}\right.\)
\(c,\Leftrightarrow\left|\dfrac{3}{4}-\dfrac{1}{2}x\right|=\dfrac{1}{2}\Leftrightarrow\left[{}\begin{matrix}\dfrac{3}{4}-\dfrac{1}{2}x=\dfrac{1}{2}\left(x\le\dfrac{3}{2}\right)\\\dfrac{1}{2}x-\dfrac{3}{4}=\dfrac{1}{2}\left(x>\dfrac{3}{2}\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\left(tm\right)\\x=\dfrac{5}{2}\left(tm\right)\end{matrix}\right.\)
\(d,\Leftrightarrow\left|5-2x\right|=4\Leftrightarrow\left[{}\begin{matrix}5-2x=4\left(x\le\dfrac{5}{2}\right)\\2x-5=4\left(x>\dfrac{5}{2}\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\left(tm\right)\\x=\dfrac{9}{2}\left(tm\right)\end{matrix}\right.\)
\(đ,\Leftrightarrow\left\{{}\begin{matrix}x-3,5=0\\x-1,3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3,5\\x=1,3\end{matrix}\right.\left(vô.lí\right)\Leftrightarrow x\in\varnothing\)
\(e,\Leftrightarrow\left\{{}\begin{matrix}x-2021=0\\x-2022=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2021\\x=2022\end{matrix}\right.\left(vô.lí\right)\Leftrightarrow x\in\varnothing\)
\(f,\Leftrightarrow\left|x\right|=\dfrac{1}{3}-x\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}-x\left(x\ge0\right)\\x=x-\dfrac{1}{3}\left(x< 0\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{6}\left(tm\right)\\0x=-\dfrac{1}{3}\left(vô.lí\right)\end{matrix}\right.\Leftrightarrow x=\dfrac{1}{6}\)
\(g,\Leftrightarrow\left[{}\begin{matrix}x-2=x\left(x\ge2\right)\\2-x=x\left(x< 2\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}0x=2\left(vô.lí\right)\\x=1\left(tm\right)\end{matrix}\right.\Leftrightarrow x=1\)
1/a, f(x) - g(x) + h(x) = x3 - 2x2 + 3x +1 - x3 - x + 1 +2x2 - 1
=(x3 - x3) + (-2x2 + 2x2) + (3x - x) + (1 + 1 - 1)
=2x + 1
b, f(x) - g(x) + h(x) = 0
<=> 2x + 1 = 0
<=> 2x = -1
<=> x = -1/2
Vậy x = -1/2 là nghiệm của đa thức f(x) - g(x) + h(x)
2/ a, 5x + 3(3x + 7)-35 = 0
<=> 5x + 9x + 21 - 35 = 0
<=> 14x - 14 = 0
<=> 14(x - 1) = 0
<=> x-1 = 0
<=> x = 1
Vậy 1 là nghiệm của đa thức 5x + 3(3x + 7) -35
b, x2 + 8x - (x2 + 7x +8) -9 =0
<=> x2 + 8x - x2 - 7x - 8 - 9 =0
<=> (x2 - x2) + (8x - 7x) + (-8 -9)
<=> x - 17 = 0
<=> x =17
Vậy 17 là nghiệm của đa thức x2 + 8x -(x2 + 7x +8) -9
3/ f(x) = g (x) <=> x3 +4x2 - 3x + 2 = x2(x + 4) + x -5
<=> x3 +4x2 - 3x + 2 = x3 + 4x2 + x - 5
<=> -3x + 2 = x - 5
<=> -3x = x - 5 - 2
<=> -3x = x - 7
<=>2x = 7
<=> x = 7/2
Vậy f(x) = g(x) <=> x = 7/2
4/ có k(-2) = m(-2)2 - 2(-2) +4 = 0
=> 4m + 4 + 4 = 0
=> 4m + 8 = 0
=> 4m = -8
=> m = -2
a; (\(x\) - 2)2.(\(x+1\)).(\(x\) - 4) < 0
(\(x-2\))2 ≥ 0 ∀\(x\); \(x+1\) = 0 ⇒ \(x=-1\); \(x-4\) = 0 ⇒ \(x=4\)
Lập bảng ta có:
\(x\) | - 1 4 |
\(x+1\) | - 0 + | + |
\(x-4\) | - | - 0 + |
(\(x-2\))2 | + | + | + |
(\(x-2\))2.(\(x+1\)).(\(x+4\)) | + 0 - 0 + |
Theo bảng trên ta có: -1 < \(x\) < 4
Vậy \(-1< x< 4\)
b; [\(x^2\).(\(x-3\)):(\(x-9\))] < 0
\(x-3=0\)⇒ \(x=3\); \(x-9\) = 0 ⇒ \(x=9\)
Lập bảng ta có:
\(x\) | 3 9 |
\(x-3\) | - 0 + | + |
\(x-9\) | - | - 0 + |
\(x^2\) | + | + | + |
\(x^2\)(\(x-3\)):(\(x-9\)) | + 0 - 0 + |
Theo bảng trên ta có: 3 < \(x\) < 9
Vậy 3 < \(x\) < 9
2:
a: 5/x-y/3=1/6
=>\(\dfrac{15-xy}{3x}=\dfrac{1}{6}\)
=>\(\dfrac{30-2xy}{6x}=\dfrac{x}{6x}\)
=>30-2xy=x
=>x(2y+1)=30
=>(x;2y+1) thuộc {(30;1); (-30;-1); (10;3); (-10;-3); (6;5); (-6;-5)}
=>(x,y) thuộc {(30;0); (-30;-1); (10;1); (-10;-2); (6;2); (-6;-3)}
b: x/6-2/y=1/30
=>\(\dfrac{xy-12}{6y}=\dfrac{1}{30}\)
=>\(\dfrac{5xy-60}{30y}=\dfrac{y}{30y}\)
=>5xy-60=y
=>y(5x-1)=60
=>(5x-1;y) thuộc {(-1;-60); (4;15); (-6;-10)}(Vì x,y là số nguyên)
=>(x,y) thuộc {(0;-60); (1;15); (-1;-10)}
a; (\(x\) - 2)2.(\(x+1\)).(\(x\) - 4) < 0
(\(x-2\))2 ≥ 0 ∀\(x\); \(x+1\) = 0 ⇒ \(x=-1\); \(x-4\) = 0 ⇒ \(x=4\)
Lập bảng ta có:
\(x\) | - 1 4 |
\(x+1\) | - 0 + | + |
\(x-4\) | - | - 0 + |
(\(x-2\))2 | + | + | + |
(\(x-2\))2.(\(x+1\)).(\(x+4\)) | + 0 - 0 + |
Theo bảng trên ta có: -1 < \(x\) < 4
Vậy \(-1< x< 4\)
b; [\(x^2\).(\(x-3\)):(\(x-9\))] < 0
\(x-3=0\)⇒ \(x=3\); \(x-9\) = 0 ⇒ \(x=9\)
Lập bảng ta có:
\(x\) | 3 9 |
\(x-3\) | - 0 + | + |
\(x-9\) | - | - 0 + |
\(x^2\) | + | + | + |
\(x^2\)(\(x-3\)):(\(x-9\)) | + 0 - 0 + |
Theo bảng trên ta có: 3 < \(x\) < 9
Vậy 3 < \(x\) < 9
a)\(\frac{x+2}{x+5}< \frac{x+1}{x+4}\)
\(\Leftrightarrow\frac{x+5-3}{x+5}< \frac{x+4-3}{x+4}\)
\(\Leftrightarrow1-\frac{3}{x+5}< 1-\frac{3}{x+4}\)
\(\Leftrightarrow\frac{3}{x+5}>\frac{3}{x+4}\)
\(\Leftrightarrow x+5< x+4\)
Vì \(x+5\)luôn lớn hơn x+4 với mọi x
nên không có giá trị x thỏa mãn
b) \(\frac{x-1}{x-2}< \frac{x+4}{x+3}\)
\(\Leftrightarrow\frac{x-2+1}{x-2}< \frac{x+3+1}{x+3}\)
\(\Leftrightarrow1+\frac{1}{x-2}< 1+\frac{1}{x+3}\)
\(\Leftrightarrow\frac{1}{x-2}< \frac{1}{x+3}\)
\(\Leftrightarrow x-2>x+3\)
Vì \(x+3>x-2\)với mọi x
nên không có giá trị x thỏa mãn