tím x biết: \(\left|2x+3\right|+\left|2x-1\right|=\dfrac{8}{3\left(x+1\right)^2+2}\)
Mn giúp mk vs ạ!!! Ai lm đúng mk sẽ tick cho
Thanks mn nhiều......
Tìm x biết :
\(\left(2x-1\right)^3+\left(3x+2\right)^3-\left(5x+1\right)^3=0\)
MẤY BẠN GIÚP MK VS Ạ AI NHANH MK VOTE NHA
tìm x thuộc Q bt:
a)\(\left(x-\dfrac{1}{2}\right)^2=0\)
b) \(\left(x+\dfrac{1}{2}\right)^2=\dfrac{1}{4}\)
c) \(\left(2.x-3\right)^3=-8\)
d) \(\left(3.x-2\right)^5=-243\)
e) \(\left(7.x+2\right)^{-1}=3^{-2}\)
f) \(\left(x-1\right)^3=-125\)
g) \(\left(2x-1\right)^4=81\)
h) \(\left(2.x-1\right)^6=\left(2.x-1\right)^8\)
các bn ai bt lm lm ơn lm phước giải ra hộ mk vs mk sẽ tik mừ .... các bn giúp mk vs nhé huhu
a)(x − 12)2 = 0
=>x − 12 = 0
=> x = 12
b) (x+12)2 = 0,25
=> x + 12 = 0,5 hoặc x + 12= -0,5
=> x = -11,5 hoặc x = -12,5
c) (2x−3)3 = -8
=> 2x - 3 = -2
=> x = 0,5
d) (3x−2)5 = −243
=> 3x - 2 = -3
=> x = -1/3
e) (7x+2)-1 = 3-2
=> \(\dfrac{1}{7x+2}=\dfrac{1}{9}\)
=> 7x + 2 = 9
=> x = 1
f) (x−1)3 = −125
=> (x−1) = −5
=> x = -4
g) (2x−1)4 = 81
=> 2x - 1 = 3
=> x = 2
h) (2x−1)6 = (2x−1)8
=> 2x -1 = 0 hoặc 2x - 1 = 1 hoặc 2x - 1 = -1
=> x = 1/2 hoặc x = 1 hoặc x = 0
a/ \(\left(x-\dfrac{1}{2}\right)^2=0\)
\(\Leftrightarrow x-\dfrac{1}{2}=0\)
\(\Leftrightarrow x=\dfrac{1}{2}\)
Vậy ...
b/ \(\left(x+\dfrac{1}{2}\right)^2=\dfrac{1}{4}\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(x+\dfrac{1}{2}\right)^2=\left(\dfrac{1}{2}\right)^2\\\left(x+\dfrac{1}{2}\right)^2=\left(-\dfrac{1}{2}\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{1}{2}\\x+\dfrac{1}{2}=-\dfrac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)
Vậy ..
c/ \(\left(2x-3\right)^3=-8\)
\(\Leftrightarrow\left(2x-3\right)^3=\left(-2\right)^3\)
\(\Leftrightarrow2x-3=-2\)
\(\Leftrightarrow x=\dfrac{1}{2}\)
Vậy ...
d/ \(\left(3x-2\right)^5=-243\)
\(\left(3x-2\right)^5=\left(-3\right)^5\)
\(\Leftrightarrow3x-2=-3\)
\(\Leftrightarrow x=-\dfrac{1}{3}\)
Vậy ...
e/ \(\left(x-1\right)^3=-125\)
\(\Leftrightarrow\left(x-1\right)^3=\left(-5\right)^3\)
\(\Leftrightarrow x-1=-5\)
\(\Leftrightarrow x=-4\)
Vậy..
f/ \(\left(2x-1\right)^4=81\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(2x-1\right)^4=3^4\\\left(2x-1\right)^4=\left(-3\right)^4\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=3\\2x-1=-3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)
Vậy...
g/ \(\left(2x-1\right)^6=\left(2x-1\right)^8\)
\(\Leftrightarrow\left(2x-1\right)^8-\left(2x-1\right)^6=0\)
\(\Leftrightarrow\left(2x-1\right)^6\left[\left(2x-1\right)^2-1\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(2x-1\right)^6=0\\\left(2x-1\right)^2-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\\left[{}\begin{matrix}2x-1=1\\2x-1=-1\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\\left[{}\begin{matrix}x=1\\x=0\end{matrix}\right.\end{matrix}\right.\)
Vậy..
MÌNH XIN LỖI, câu A và B mình ghi sai đề
a) \(\left(x-\dfrac{1}{2}\right)^2=0\)
\(\Rightarrow x-\dfrac{1}{2}=0\)
\(\Rightarrow x=\dfrac{1}{2}\)
b)\(\left(x+\dfrac{1}{2}\right)^2=\dfrac{1}{4}\)
\(\Rightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{1}{2}\\x+\dfrac{1}{2}=-\dfrac{1}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)
\(\sqrt{\left(1+2x\right)\left(3-x\right)}=2x^2-5x+3+m\)
Tìm m để pt có nghiệm
mn giúp mk vs ạ
Rút gọn các biểu thức sau:
a/\(\left(x+\dfrac{1}{3}x+\dfrac{1}{9}\right)\left(x-\dfrac{1}{3}\right)-\left(x-\dfrac{1}{3^{ }}\right)^2\)
b/\(\left(x_{ }^2-2\right)^3-x\left(x+1\right)\left(x-1\right)+x\left(x-3\right)\)
MẤY BẠN GIÚP MK VS Ạ AI NHANH MK VOTE NHA
a) \(=x^3-\dfrac{1}{27}-x^2+\dfrac{2}{3}x-\dfrac{1}{9}=x^3-x^2+\dfrac{2}{3}x-\dfrac{2}{27}\)
b) \(=x^6-6x^4+12x^2-8-x^3+x+x^2-3x=x^6-6x^4-x^3+13x^2-2x-8\)
Tìm x:
\(3.\left(2x-6\right)-4.\left(1+2x\right)-2.\left(x-4\right)=4-3\left(1+2x\right)-5.\left(1-2x\right)\)
Ai nhanh mk tick cho . Mk cảm ơn bạn nhiều .
\(3\left(2x-6\right)-4\left(1+2x\right)-2\left(x-4\right)=4-3\left(1+2x\right)-5\left(1-2x\right).\)
\(\Leftrightarrow6x-18-4-8x-2x+8=4-3-6x-5+10x\)
\(\Leftrightarrow-4x-14=4x-4\)
\(\Leftrightarrow-4x-4x=-4+14\)
\(\Leftrightarrow-8x=10\)
\(\Leftrightarrow x=-\frac{5}{4}\)
tìm giá trị nhỏ nhất
e) E= \(2.\left|x-\dfrac{1}{2}\right|+2021\)
g) G= \(\left|x-1\right|+\left|x-2\right|\)
h) H= \(\left|x-1\right|+\left|x-2\right|+\left|x-3\right|\)
k) K= \(\left|x-1\right|+\left|2x-1\right|\)
lm nhanh giúp mk nhé mk đang cần gấp lắm
1
e) E >= 2021
dấu = xảy ra khi x=1/2
g) G = |x-1|+ |2-x| >= |x-1+2-x|=1
Dấu = xảy ra khi (x-1)(2-x)>=0 <=> 1<=x<=2
h) H = |x-1|+|x-2| + |x-3|
Ta có : |x-1| + |x-3| = |x-1| + |3-x| >= |x-1+3-x| = 2
|x-2| >=0
=> H>=2
Dấu = xảy ra khi (x-1)(3-x) >=0 ; x-2=0
<=> x=2
k) K = |x-1| + |2x-1|
2K = |2x-2| + |2x-1| + |2x-1|
Ta có : |2x-2| + |2x-1| = |2x-2| + |1-2x| >= |2x-2+1-2x|=1
|2x-1| >=0
Dấu = xảy ra (2x-2)(1-2x) >=0; 2x-1=0
<=> x=1/2
e)Vì \(\left|x-\dfrac{1}{2}\right|\ge0\forall x\)
\(\Leftrightarrow2\left|x-\dfrac{1}{2}\right|\ge0\forall x\\ \Rightarrow2\left|x-\dfrac{1}{2}\right|+2012\ge2012\forall x\)
Dấu "=" xảy ra khi x=\(\dfrac{1}{2}\)
Vậy...
b)G=|x-1|+ |2-x|\(\)
áp dụng bđt |a+b|+ |c+d|\(\ge\left|a+b+c+d\right|\forall x\)
\(\Rightarrow\)ta có |x-1|+ |2-x|\(\ge\) \(\left|x-1+2-x\right|\forall x\)
\(\Leftrightarrow\text{|x-1|+ |2-x| }\ge1\forall x\)
Dấu "=" xảy ra khi 1\(\le x\le2\) \(\forall x\)
Vậy...
h)H= |x-1|+|x-2| + |x-3|
Ta có |x-1| + |x-3|
=|x-1| + |3-x| ( trong giá trị tuyệt đối đổi dấu không cần đặt dấu trừ ở ngoài)
=>|x-1| + |3-x|\(\ge\left|x-1+3-x\right|\forall x\)
<=>|x-1| + |3-x|\(\ge2\forall x\) (1)
Mà |x-2|\(\ge0\forall x\) (2)
Từ (1) và (2)=> ta có |x-1|+|x-2| + |x-3| \(\ge2\forall x\)
Dấu "=" xảy ra khi x-2=0
<=>x=2
Vậy...
k) K = |x-1| + |2x-1|
2K = |2x-2| + |2x-1| + |2x-1|
Mà : |2x-2| + |2x-1|
=|2x-2| + |1-2x|\(\ge\text{|2x-2+1-2x|}\) \(\forall x\)
Lại có |2x-1| \(\ge\)0 \(\forall x\)
Dấu "=" xảy ra 2x-1=0
<=>x=\(\dfrac{1}{2}\)
Vậy....
Tìm giá trị nhỏ nhất
e) E=\(2.\left|x-\dfrac{1}{2}\right|+2021\)
g) G=\(\left|x-1\right|+\left|x-2\right|\)
h) H=\(\left|x-1\right|+\left|x-2\right|+\left|x-3\right|\)
k) K=\(\left|x-1\right|+\left|2x-1\right|\)
Lm nhanh giúp mk nhé!Mk đang cần gấp lắm
tìm giá trị nhỏ nhất
e)E= \(2.\left|x-\dfrac{1}{2}\right|+2021\)
g) G=\(\left|x-1\right|+\left|x-2\right|\)
h) H=\(\left|x-1\right|+\left|x-2\right|+\left|x-3\right|\)
k) K=\(\left|x-1\right|+\left|2x-1\right|\)
lm nhanh giúp mk nhé! mk đang cần gấp lắm
e) Ta có: \(2\left|x-\dfrac{1}{2}\right|\ge0\forall x\)
\(\Leftrightarrow2\left|x-\dfrac{1}{2}\right|+2021\ge2021\forall x\)
Dấu '=' xảy ra khi \(x=\dfrac{1}{2}\)
\(d)\left|x+1\right|+\left|x+2\right|+\left|2x-3\right|=6\)
\(e)\left|x+1\right|+\left|x+2\right|+\left|x-3\right|+\left|x-5\right|=7\)
mình đag cần gấp. ai lm nhanh mình tick. mong mn giúp mình
Ta có bất đẳng thức giá trị tuyệt đối:
\(\left|A\right|+\left|B\right|\ge\left|A+B\right|\)
Dấu \(=\)khi \(AB\ge0\).
d) \(\left|x+1\right|+\left|x+2\right|+\left|2x-3\right|\)
\(\ge\left|x+1+x+2\right|+\left|2x-3\right|\)
\(=\left|2x+3\right|+\left|3-2x\right|\)
\(\ge\left|2x+3+3-2x\right|=6\)
Dấu \(=\)khi \(\hept{\begin{cases}\left(x+1\right)\left(x+2\right)\ge0\\\left(2x+3\right)\left(3-2x\right)\ge0\end{cases}}\Leftrightarrow-1\le x\le\frac{3}{2}\).
e) \(\left|x+1\right|+\left|x+2\right|+\left|x-3\right|+\left|x-5\right|\)
\(=\left(\left|x+1\right|+\left|3-x\right|\right)+\left(\left|x+2\right|+\left|5-x\right|\right)\)
\(\ge\left|x+1+3-x\right|+\left|x+2+5-x\right|\)
\(=4+7=11\)
Dấu \(=\)khi \(\hept{\begin{cases}\left(x+1\right)\left(3-x\right)\ge0\\\left(x+2\right)\left(5-x\right)\ge0\end{cases}}\Leftrightarrow-1\le x\le3\).
Do đó phương trình đã cho vô nghiệm.