tìm x
a, I 2x + 1 I = 7
b, 3 I x + 1 I + 1 = 28
c, I 7x - 1 I + 8 = 7
d, I 2x + 1 I = I x - 1 I
cần gấp lắm nha mn, tick 3 ngày
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\(\left(x-1\right)^3-2\left(x+1\right)^2=\left(2x+1\right)\left(1-3x\right)-2x\left(1-x\right)\)
\(\Leftrightarrow x^3-3x^2+3x-1-2x^2-4x-2=2x-6x^2+1-3x-2x+2x^2\)
\(\Leftrightarrow x^3-5x^2-x-3=-4x^2-3x+1\)
\(\Leftrightarrow x^3+x^2+2x-4=0\)
\(\Leftrightarrow x^3-x^2+2x^2-2x+4x-4=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2+2x+4\right)=0\)
=>x-1=0
hay x=1
\(1)\) Ta có :
\(\left|2x-1\right|\ge0\)
\(\Leftrightarrow\)\(A=\left|2x-1\right|+8\ge8\)
Dấu "=" xảy ra \(\Leftrightarrow\)\(\left|2x-1\right|=0\)
\(\Leftrightarrow\)\(2x-1=0\)
\(\Leftrightarrow\)\(2x=1\)
\(\Leftrightarrow\)\(x=\frac{1}{2}\)
Vậy GTNN của \(A\) là \(8\) khi \(x=\frac{1}{2}\)
Chúc bạn học tốt ~
\(2)\) Ta có :
\(B=\left|x-3\right|+\left|x-9\right|-1\)
\(B=\left|x-3\right|+\left|9-x\right|-1\ge\left|x-3+9-x\right|-1=\left|6\right|-1=6-1=5\)
Dấu "=" xảy ra khi và chỉ khi \(\left(x-3\right)\left(9-x\right)\ge0\)
Trường hợp 1 :
\(\hept{\begin{cases}x-3\ge0\\9-x\ge0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ge3\\x\le9\end{cases}\Leftrightarrow}3\le x\le9}\)
Trường hợp 2 :
\(\hept{\begin{cases}x-3\le0\\9-x\le0\end{cases}\Leftrightarrow\hept{\begin{cases}x\le3\\x\ge9\end{cases}}}\) ( loại )
Vậy GTNN của \(B\) là \(5\) khi \(3\le x\le9\)
Chúc bạn học tốt ~
\(1a,\frac{\left(2x+1\right)^2}{5}-\frac{\left(x-1\right)^2}{3}=\frac{7x^2-14x-5}{15}\)
\(\Leftrightarrow\frac{3\left(2x+1\right)^2}{15}-\frac{5\left(x-1\right)^2}{15}=\frac{7x^2-14x-5}{15}\)
\(\Leftrightarrow\frac{12x^2+12x+3}{15}-\frac{5x^2-10x+5}{15}=\frac{7x^2-14x-5}{15}\)
\(\Leftrightarrow12x^2+12x+3-5x^2+10x-5=7x^2-14x-5\)
\(\Leftrightarrow36x=-3\)
\(x=-\frac{1}{12}\)
Vậy ................
\(b,\frac{7x-1}{6}+2x=\frac{16-x}{5}\)
\(\Leftrightarrow\frac{5\left(7x-1\right)}{30}+\frac{30.2x}{30}=\frac{6\left(16-x\right)}{30}\)
\(\Leftrightarrow35x-5+60x=96-6x\)
\(\Leftrightarrow101x=101\)
\(\Leftrightarrow x=1\)
Vậy ....................
\(\left|5-7x\right|=\dfrac{1}{4}\)\(\Rightarrow\left\{{}\begin{matrix}5-7x=\dfrac{1}{4}\\5-7x=\dfrac{-1}{4}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}7x=\dfrac{19}{4}\\7x=\dfrac{21}{4}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{19}{28}\\x=\dfrac{3}{4}\end{matrix}\right.\)\(\left|4x-11\right|=\dfrac{1}{2}x-1\left\{{}\begin{matrix}4x-11=\dfrac{1}{2}x-1\\4x-11=-\left(\dfrac{1}{2}x-1\right)\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}4x-\dfrac{1}{2}x=11-1\\4x-11=-\dfrac{1}{2}x+1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x\left(4-\dfrac{1}{2}\right)=10\\4x+\dfrac{1}{2}x=11+1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x\times\dfrac{7}{2}=10\\x\left(4+\dfrac{1}{2}\right)=12\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{20}{7}\\x\times\dfrac{9}{2}=12\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{20}{7}\\x=\dfrac{8}{3}\end{matrix}\right.\)
a/ \(\left|\frac{3x-6}{1-2x}\right|=x-2\) \(\left(x\ne\frac{1}{2}\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}\frac{3x-6}{1-2x}=x-2\\\frac{3x-6}{1-2x}=2-x\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}3x-6=\left(x-2\right)\left(1-2x\right)\\3x-6=\left(2-x\right)\left(1-2x\right)\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}3x-6=x+4x-2-2x^2\\3x-6=-x-4x+2+2x^2\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}-2x^2+2x+4=0\\2x^2-8x+8=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\\x=2\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)
KL: .............
b/ Tương tự
a ) \(A=\left|x+1\right|+\left|x+2\right|-2x+3\ge2x+3-2x+3=6\)
Dấu " = " xảy ra khi \(\left(x+2\right)\left(x+1\right)\ge0\)
b )
\(B=\left|2x+3\right|+\left|1-2x\right|\ge\left|2x+3+1-2x\right|=4\)
Dấu " = " xảy ra khi \(\left(2x+3\right)\left(1-2x\right)\ge0\)
c )
\(C=\left|x-1\right|+\left|x-2\right|+\left|x-2\right|\ge\left|x-1\right|+\left|2-x\right|\ge\left|x-1+2-x\right|=1\)
Dấu " = " xảy ra khi \(x=2\)
a) |2x +1| = 7
Th1: 2x + 1 = 7
<=> x = 3
Th2: 2x + 1 = -7
<=> x = -4