Tìm x : (x - 1) ^ x + 2 = (x - 1) ^ x + 6 với x ∈ Z
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Tìm x thuộc Z: (x-1)x+2=(x-1)x+6.
Giúp em với ạ sáng mai em phải nộp bài rồi ạ
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Hai luỹ thừa có cơ số giống nhau, số mũ khác nhau mà bằng nhau => cơ số là 1 hoặc -1
Ta có:
x - 1 = 1
x = 1 + 1
x = 2
Hoặc
x - 1 = -1
x = -1 + 1
x = 0
Vậy, x = 1 hoặc 0
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\(x = {x +y-1\over z}= {y+z-1\over x}= {z+x+2\over y}\)
với x,y,z không bằng 0. Tìm x,y,z
các đại nhân giúp em với ạ
Nhờ mọi người giải giúp em hai bài toán này với ạ .
1) giải phương trình :
x +3x/√(x^2-9) =6√2
1) Cho các số thực dương thỏa mãn √(x^2+y^2) +√(y^2+z^2) +√(z^2+x^2) = 2015
Tìm giá trị nhỏ nhất của T=x^2/(y+z) +y^2/(z+x) +z^2/(x+y)
Tìm x biết :
a)2/3(x-1)-x-3/4=1
b)5/6(x+2)-x-1/2=1/3
Mn ơi giúp em với ạ. Em cần gấp ạ
\(\frac{2}{3}\left(x-1\right)-x-\frac{3}{4}=1\)
<=> \(\frac{2}{3}x-\frac{2}{3}-x-\frac{3}{4}=1\)
<=> \(-\frac{1}{3}x-\frac{17}{12}=1\)
<=> \(-\frac{1}{3}x=\frac{29}{12}\)
<=> \(x=-\frac{29}{4}\)
\(\frac{5}{6}\left(x+2\right)-x-\frac{1}{2}=\frac{1}{3}\)
<=> \(\frac{5}{6}x+\frac{5}{3}-x-\frac{1}{2}=\frac{1}{3}\)
<=> \(-\frac{1}{6}x+\frac{7}{6}=\frac{1}{3}\)
<=> \(-\frac{1}{6}x=-\frac{5}{6}\)
<=> \(x=5\)
học tốt
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Tìm giá trị nhỏ nhất của biểu thức B = (x – 1)(x + 2)(x + 3)(x + 6)
giúp em với ạ ;-;
B=\(\left(x-1\right)\left(x+2\right)\left(x+3\right)\left(x+6\right)\)=\(\left(x^2+5x-6\right)\left(x^2+5x+6\right)\)
- Đặt t=\(x^2+5x-6\)
=>B=t(t+12)=t2+12t=(t2+12t+36)-36 =(t+6)2-36≥-36
- minB=-36 ⇔ t+6=0 ⇔\(x^2+5x-6+6=0\) ⇔\(x\left(x+5\right)=0\) ⇔x=0 hay x=-5.
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1)Tính hợp lý:
a: (36-16).(-5)+6.(-14-6)
2)Tìm x thuộc Z
a/ x+(-23)=(-100)+77
b/ x-(5-x)=x-15
GIÚP MÌNH VỚI MÌNH CẦN GẤP Ạ, MÌNH CẢM ƠN NHIỀU
2. a) x + (-23) = (-100) + 77
1)
a: (36-16).(-5)+6.(-14-6)
=20.(-5)+6.(-20)
=20.(-5)+(-6).20
=20.(-5+-6)
=20.(-11)
=-220
2)
a/x+(-23)=(-100)+77
x+(-23)=-23
x=(-23)+(-23)
x=0
Vậy x=0
tìm phần số x/9 (x thuộc z) sao cho:
x/9<4/7<x+1/9
GIÚP MÌNH VỚI Ạ. MÌNH CẦN GẤP. CẢM ƠN CÁC BẠN NHIỀU!
CÔ NGUYỄN THỊ THƯƠNG HOÀI GIÚP EM VỚI Ạ
\(\dfrac{x}{9}\) < \(\dfrac{4}{7}\) < \(x\) + \(\dfrac{1}{9}\)
\(\dfrac{7x}{63}\) < \(\dfrac{36}{63}\) < \(\dfrac{63x}{63}\) + \(\dfrac{7}{63}\)
7\(x\) < 36 < 63\(x\) + 7
⇒\(\left\{{}\begin{matrix}7x< 36\\63x+7>36\end{matrix}\right.\)⇒\(\left\{{}\begin{matrix}x< \dfrac{36}{7}\\63x>36-7\end{matrix}\right.\)⇒\(\left\{{}\begin{matrix}x< \dfrac{36}{7}\\63x>29\end{matrix}\right.\)⇒\(\left\{{}\begin{matrix}x< \dfrac{36}{7}\\x>\dfrac{29}{63}\end{matrix}\right.\)
\(\dfrac{29}{63}\)< \(x\) < \(\dfrac{36}{7}\) vì \(x\in\) Z nên \(x\in\) { 1; 2; 3; 4; 5}
⇒ \(\dfrac{x}{9}\) = \(\dfrac{1}{9}\); \(\dfrac{2}{9}\); \(\dfrac{3}{9}\); \(\dfrac{4}{9}\);\(\dfrac{5}{9}\)
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tìm phần số x/9 (x thuộc z) sao cho:
x/9<4/7<x+1/9
GIÚP MÌNH VỚI Ạ. MÌNH CẦN GẤP. CẢM ƠN CÁC BẠN NHIỀU!
CÔ NGUYỄN THỊ THƯƠNG HOÀI GIÚP EM VỚI Ạ
\(\dfrac{x}{9}< \dfrac{4}{7}< \dfrac{x+1}{9}\)
=>\(\dfrac{7x}{63}< \dfrac{36}{63}< \dfrac{7x+7}{63}\)
\(\Rightarrow7x< 36< 7x+7\)
\(\Rightarrow x< \dfrac{36}{7}< x+1\)
\(\Rightarrow x< 5\dfrac{1}{7}< x+1\)
\(\Rightarrow x=5\)
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cho các số dương x y z thỏa mãn x+y+z=2
Tìm min P = \(\dfrac{x^2}{y+z}\)+\(\dfrac{y^2}{z+x}\)+\(\dfrac{z^2}{x+y}\)
Thầy Lâm giúp với em với ạ
Lời giải:
Áp dụng BĐT AM-GM:
$\frac{x^2}{y+z}+\frac{y+z}{4}\geq 2\sqrt{\frac{x^2}{y+z}.\frac{y+z}{4}}=x$
$\frac{y^2}{x+z}+\frac{x+z}{4}\geq y$
$\frac{z^2}{x+y}+\frac{x+y}{4}\geq z$
Cộng theo vế các BĐT trên và thu gọn ta được:
$P\geq \frac{x+y+z}{2}=\frac{2}{2}=1$
Vậy $P_{\min}=1$ khi $x=y=z=\frac{2}{3}$
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