Tìm x biết
a,|x + 1| + |x + 2| + |x + 3| = 2x
b, |x + 1| + |x + 2| = x
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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) 7/x-1=x+1/9
7/x-x-1=1/9
7/x-x=1/9+1
7/x-x=10/9
còn lại thì bạn tự tính nhé
câu b làm tương tự
b) \(x^3-x^2-x+1=0\Leftrightarrow\left(x-1\right)^2\left(x+1\right)=0\)
\(\Leftrightarrow x-1=0\) hoặc \(x+1=0\)
\(\Leftrightarrow x=1\) hoặc \(x=-1\)
c) \(x^2-6x+8=0\Leftrightarrow\left(x-4\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=2\end{matrix}\right.\)
a) \(x^3+x^2+x+1=0\Leftrightarrow\left(x+1\right)\left(x^2+1\right)=0\Leftrightarrow x+1=0\Leftrightarrow x=-1\)
(do \(x^2+1\ge1>0\))
Lời giải:
a. $2x^2+3(x-1)(x+1)=5x(x+1)$
$\Leftrightarrow 2x^2+3x^2-3=5x^2+5x$
$\Leftrightarrow 5x^2-3=5x^2+5x$
$\Leftrightarrow 5x=-3$
$\Leftrightarrow x=\frac{-3}{5}$
b.
PT $\Leftrightarrow (-5x^2-2x+16)+4(x^2-x-2)=4-x^2$
$\Leftrightarrow -x^2-6x+8=4-x^2$
$\Leftrightarrow -6x+8=4$
$\Leftrightarrow -6x=-4$
$\Leftrightarrow x=\frac{2}{3}$
c.
PT $\Leftrightarrow 4(x^2+4x-5)-(x^2+7x+10)=3(x^2+x-2)$
$\Leftrightarrow 4x^2+16x-20-x^2-7x-10=3x^2+3x-6$
$\Leftrightarrow 3x^2+9x-30=3x^2+3x-6$
$\Leftrightarrow 6x=24$
$\Leftrightarrow x=4$
a: Ta có: \(\left(2x+1\right)^2-4\left(x+2\right)^2=9\)
\(\Leftrightarrow4x^2+4x+1-4x^2-16x-16=9\)
\(\Leftrightarrow-12x=24\)
hay x=-2
b: Ta có: \(\left(x+3\right)^2-\left(x-4\right)\left(x+8\right)=1\)
\(\Leftrightarrow x^2+6x+9-x^2-4x+32=1\)
\(\Leftrightarrow2x=-40\)
hay x=-20
\(a,\Rightarrow x^2+4x+4+x^2-2x+1+x^2-9-3x^2=-8\\ \Rightarrow2x=-4\Rightarrow x=-2\\ b,\Rightarrow\left(x-2021\right)\left(2022x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=2021\\x=\dfrac{1}{2022}\end{matrix}\right.\\ c,\Rightarrow\left(x^2-9\right)-\left(x-3\right)\left(2x+7\right)=0\\ \Rightarrow\left(x-3\right)\left(x+3\right)-\left(x-3\right)\left(2x+7\right)=0\\ \Rightarrow\left(x-3\right)\left(x+3-2x-7\right)=0\\ \Rightarrow\left(x-3\right)\left(-4-2x\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
Lời giải:
a. Đề có cả x,y. Bạn xem lại
b.
PT $\Leftrightarrow 5x(x-3)-2(x-3)=0$
$\Leftrightarrow (x-3)(5x-2)=0$
$\Leftrightarrow x-3=0$ hoặc $5x-2=0$
$\Leftrightarrow x=3$ hoặc $x=\frac{2}{5}$
c.
PT $\Leftrightarrow (7x-2)(x-4)=0$
$\Leftrightarrow 7x-2=0$ hoặc $x-4=0$
$\Leftrightarrow x=\frac{2}{7}$ hoặc $x=4$
d. Đề thiếu.
\(\left(x+1\right)\left(x+3\right)-\left(x-2\right)\left(x+5\right)=2\)
\(\Leftrightarrow x^2+4x+3-x^2-3x+10=2\)
\(\Leftrightarrow x=-11\)
`#3107.101107`
`1.`
`a,`
`(2x - 3)^2 = |3 - 2x|`
`=> (2x - 3)^2 = |2x - 3|`
`=>`\(\left[{}\begin{matrix}2x-3=\left(2x-3\right)^2\\2x-3=-\left(2x-3\right)^2\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}2x-3-\left(2x-3\right)^2=0\\2x-3+\left(2x-3\right)^2=0\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}\left(2x-3\right)\left(1-2x+3\right)=0\\\left(2x-3\right)\left(1+2x-3\right)=0\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}2x-3=0\\4-2x=0\\2x-2=0\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=2\\x=1\end{matrix}\right.\)
Vậy, `x \in {3/2; 2; 1}`
`b,`
`(x - 1)^2 + (2x - 1)^2 = 0`
`=>`\(\left[{}\begin{matrix}\left(x-1\right)^2=0\\\left(2x-1\right)^2=0\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x-1=0\\2x-1=0\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=1\\x=\dfrac{1}{2}\end{matrix}\right.\)
Vậy, `x \in {1; 1/2}`
`c,`
`5 - x^2 = 1`
`=> x^2 = 4`
`=> x^2 = (+-2)^2`
`=> x = +-2`
Vậy, `x \in {-2; 2}`
`d,`
`x - 2\sqrt{x} = 0`
`=> x^2 - (2\sqrt{x})^2 = 0`
`=> x^2 - 4x = 0`
`=> x(x - 4) = 0`
`=>`\(\left[{}\begin{matrix}x=0\\x-4=0\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)
Vậy, `x \in {0; 4}`
`g,`
`(x - 1) + 1/7 = 0`
`=> x - 1 + 1/7 = 0`
`=> x - 6/7 = 0`
`=> x = 6/7`
Vậy, `x = 6/7.`
a,Vì \(\left|x+1\right|\ge0\forall x\in R\)
\(\left|x+2\right|\) .................
\(\left|x+3\right|\) .................
\(\Rightarrow\left|x+1\right|+\left|x+2\right|+\left|x+3\right|\ge0\forall x\in R\)
=> \(2x\ge0\forall x\in R\Leftrightarrow x\ge0\)
\(\Rightarrow x+1+x+2+x+3=2x\)
\(3x+6=2x\)
\(\Rightarrow x=-6\)
Vậy.............
b) Vì \(\left|x+1\right|\ge0\forall x\in R\)
\(\left|x+2\right|\ge0\forall x\in R\)
\(\Rightarrow\left|x+1\right|+\left|x+2\right|\ge0\forall x\in R\)
=> \(x\ge0\)
\(\Rightarrow x+1+x+2=x\)
\(2x+3=x\)
\(x=-3\)
Vậy x=3
hc tốt