Tìm x biết: \(\left|1-2x\right|=\left|3+2x\right|\)
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10 tháng 3 2022 lúc 20:28
tìm x , biết
\(a,\left|2x-3\right|-\left|3x+2\right|=0\\ b,\left|\dfrac{1}{2}x\right|=3-2x\)
a)TH1: \(2x-3>0;3x+2>0\)
\(=>2x-3-3x-2=0\\ =>-x-5=0\\ =>-x=5=>x=-5\)
TH2: \(2x-3< 0;3x+2< 0\)
\(=>-2x+3+3x+2=0\\ =>x+5=0\\ =>x=-5\)
Cả 2 TH ra \(x=-5=>x=-5\)
b)TH1 \(\dfrac{1}{2}x>0\)
\(=>\dfrac{1}{2}x=3-2x\\ =>3-2x-\dfrac{1}{2}x=0\\ =>\dfrac{4}{2}x-\dfrac{1}{2}x=3\\ =>\dfrac{3}{2}x=3\\ =>x=2\)
TH2 \(\dfrac{1}{2}x< 0\)
\(=>-\dfrac{1}{2}x=3-2x\\ =>3-2x+\dfrac{1}{2}x=0\\ =>\dfrac{4}{2}x+\dfrac{1}{2}x=3\\ =>\dfrac{5}{2}x=3\\ =>x=\dfrac{6}{5}\)
\(=>x=2;\dfrac{6}{5}\)
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BÀI 6 tìm x1,2xleft(x-5right)-left(3x+2x^2right)0 2,xleft(5-2xright)+2xleft(x-1right)133,2x^3left(2x-3right)-x^2left(4x^2-6x+2right)0 4,5xleft(x-1right)-left(x+2right)left(5x-7right)65,6x^2-left(2x-3right)left(3x+2right)1 6,2xleft(1-xright)+59-2x^2
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BÀI 6 tìm x
1,\(2x\left(x-5\right)-\left(3x+2x^2\right)=0\) 2,\(x\left(5-2x\right)+2x\left(x-1\right)=13\)
3,\(2x^3\left(2x-3\right)-x^2\left(4x^2-6x+2\right)=0\) 4,\(5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\)
5,\(6x^2-\left(2x-3\right)\left(3x+2\right)=1\) 6,\(2x\left(1-x\right)+5=9-2x^2\)
1: \(\Leftrightarrow2x^2-10x-3x-2x^2=0\)
=>-13x=0
=>x=0
2: \(\Leftrightarrow5x-2x^2+2x^2-2x=13\)
=>3x=13
=>x=13/3
3: \(\Leftrightarrow4x^4-6x^3-4x^3+6x^3-2x^2=0\)
=>-2x^2=0
=>x=0
4: \(\Leftrightarrow5x^2-5x-5x^2+7x-10x+14=6\)
=>-8x=6-14=-8
=>x=1
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`1)2x(x-5)-(3x+2x^2)=0`
`<=>2x^2-10x-3x-2x^2=0`
`<=>-13x=0`
`<=>x=0`
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`2)x(5-2x)+2x(x-1)=13`
`<=>5x-2x^2+2x^2-2x=13`
`<=>3x=13<=>x=13/3`
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`3)2x^3(2x-3)-x^2(4x^2-6x+2)=0`
`<=>4x^4-6x^3-4x^4+6x^3-2x^2=0`
`<=>x=0`
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`4)5x(x-1)-(x+2)(5x-7)=0`
`<=>5x^2-5x-5x^2+7x-10x+14=0`
`<=>-8x=-14`
`<=>x=7/4`
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`5)6x^2-(2x-3)(3x+2)=1`
`<=>6x^2-6x^2-4x+9x+6=1`
`<=>5x=-5<=>x=-1`
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`6)2x(1-x)+5=9-2x^2`
`<=>2x-2x^2+5=9-2x^2`
`<=>2x=4<=>x=2`
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tìm x, biết
\(3\left(2x-1\right)\left(3x-1\right)-\left(2x-3\right)\left(9x-1\right)=0\)
1,tìm x biết:\(\left|2x+3\right|+\left|2x-1\right|=\dfrac{8}{3.\left(x+1\right)^2+2}\)
\(VT=\left|2x+3\right|+\left|2x-1\right|=\left|2x+3\right|+\left|1-2x\right|\ge\left|2x+3+1-2x\right|=\left|4\right|=4\)
\(VP=\frac{8}{3\left(x+1\right)^2+2}\le\frac{8}{2}=4\)
\(VT\ge VP\)
Dấu "=" xảy ra \(\Leftrightarrow\)\(\hept{\begin{cases}\left(2x+3\right)\left(1-2x\right)\ge0\left(1\right)\\\left(x+1\right)^2=0\left(2\right)\end{cases}}\)
\(\left(2\right)\)\(\Leftrightarrow\)\(x=-1\) ( thỏa mãn\(\left(1\right)\) )
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a) Tìm số tự nhiên n biết \(\left(n-1\right)^{n+11}-\left(n-1\right)^n=0\)
b) Tìm x biết: \(3\left(x-2\right)-4\left(2x+1\right)-5\left(2x+3\right)=50\)
c) Tìm x biết: \(\left|2x-3\right|=\left|2-x\right|\)
b) 3x - 6 - (8x + 4) - (10x + 15) = 50
=> 3x - 6 - 8x - 4 - 10x - 15 = 50
=> (3x - 8x - 10x) = 6+ 4 + 15 + 50
=> -15x = 75 => x = 75 : (-15) = -5
c) => 2x - 3 = 2 - x hoặc 2x - 3 = - (2 - x) (Vì 2 số có giá trị tuyệt đối bằng nhau thì chings bằng nhau hoặc đối nhau)
+) nếu 2x - 3 = 2 - x => 2x+ x = 2 + 3 => 3x = 5 => x = 5/3
+) nếu 2x - 3 = -(2 - x) => 2x - 3 = -2 + x => 2x - x = -2 + 3 => x = 1
Vậy x = 5/3 hoặc x = 1
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a) (n-1)n+11-(n-1)n=0
(n-1)n(n-1)11-(n-1)n=0
(n-1)n[(n-1)11-1]=0
(n-1)n=0 hoặc (n-1)11-1=0
n-1=0 hoặc (n-1)11 =1
n=1 hoặc n-1 =1
n=1 hoặc n =2
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Tìm x biết \(\left|5\left(2x+3\right)\right|+\left|2\left(2x+3\right)\right|+\left|2x+3\right|=16\)
NX: 2x+3; 5(2x+3) và 2(2x+3) cùng dấu
+TH1: 2x+3 \(\ge\)0 => x \(\ge\frac{-3}{2}\)
=> 5(2x+3), 2(2x+3) \(\ge\)0
=> |5(2x+3)| = 5(2x+3); |2(2x+3)| = 2(2x+3); |2x+3| = 2x+3
=> (2x+3)(5+2+1) = 16
=> 2x+3 = 2
=> 2x = -1
=> x = -1/2 (t/m)
+ TH2: 2x+3 < 0 => x < -3/2
cmtt => -5(2x+3) - 2(2x+3) - (2x+3) = 16
=> (2x+3)(-5-2-1) = 16
=> 2x+3 = -2
=> 2x = -5
=> x = -5/2 (t/m)
/8(2x+3/ = 16
/2x+3/=2
2x+3=2 hoặc 2x+3=-2
2x=-1 hoặc 2x=-5
x=-1/2 hoặc x=-5/2
bạn trả lời nhé
Tìm x biết:
a) \(\left(2x-3\right).\left(3-x\right)\le0\)
b) \(\left(2x-3\right).\left(1-2x\right)>0\)
a) \(\left(2x-3\right).\left(3-x\right)\le0\)
Xét 2 trường hợp:
TH1: \(\begin{cases}2x-3\le0\\3-x\ge0\end{cases}\)\(\Rightarrow\begin{cases}2x\le3\\3\ge x\end{cases}\)\(\Rightarrow\begin{cases}x\le\frac{3}{2}\\x\le3\end{cases}\)\(\Rightarrow x\le\frac{3}{2}\)TH2: \(\begin{cases}2x-3\ge0\\3-x\le0\end{cases}\)\(\Rightarrow\begin{cases}2x\ge3\\3\le x\end{cases}\)\(\Rightarrow\begin{cases}x\ge\frac{3}{2}\\x\ge3\end{cases}\)\(\Rightarrow x\ge3\)Vậy \(\left[\begin{array}{nghiempt}x\le\frac{3}{2}\\x\ge3\end{array}\right.\) thỏa mãn đề bài
b) (2x - 3).(1 - 2x) > 0
=> 2x - 3 và 1 - 2x là 2 số cùng dấu
Xét 2 trường hợp
TH1: \(\begin{cases}2x-3< 0\\1-2x< 0\end{cases}\)\(\Rightarrow\begin{cases}2x< 3\\1< 2x\end{cases}\)\(\Rightarrow\begin{cases}x< \frac{3}{2}\\\frac{1}{2}< x\end{cases}\)\(\Rightarrow\frac{1}{2}< x< \frac{3}{2}\), thỏa mãnTH2: \(\begin{cases}2x-3>0\\1-2x>0\end{cases}\)\(\Rightarrow\begin{cases}2x>3\\1>2x\end{cases}\)\(\Rightarrow\begin{cases}x>\frac{3}{2}\\\frac{1}{2}>x\end{cases}\)\(\Rightarrow\frac{1}{2}>x>\frac{3}{2}\), vô lýVẫy \(\frac{1}{2}< x< \frac{3}{2}\) thỏa mãn đề bài
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tìm x biết
a.\(\left(x-2\right)^3-\left(x-3\right)\left(x^2+3x+9\right)6\left(x+1\right)^2=49\)49
b.\(\left(x+2\right)\left(x^2-2x+4\right)-x\left(x^2+2\right)=25\)
c.\(\left(x+3\right)^3-x\left(3x+1\right)^2+\left(2x+1\right)\left(4x^2-2x+1\right)=28\)
Tìm x biết
a) \(\left(x+3\right)^3-x\left(3x+1\right)^2+\left(2x+1\right)\left(4x^2-2x+1\right)-3x^2=54\)
Tìm x,biết
\(\left(x+3\right)^3-x\left(3x+1\right)^2+\left(2x+1\right)\left(4x^2-2x+1\right)-3x^2=42\) 42