Xem lại đề x + 3 hay x + 2 

Giá tiền phải trả khi mua xoài là: 

`25000x` (đồng)

Giá tiền phải trả khi mua nhãn là:

`20000y` (đồng)

Mà tổng số tiền phải trả là 200000 đồng nên ta có pt:

`25000x+20000y=200000`

`<=>25x+20y=200`

`<=>5x+4y=40`

2 Nghiệm của pt là: (4;5); (0;10) 

**I'm planning to buy a new house. I 1) have been looking (look) for one for two months now. So far I 2) have looked (look) at ten houses, but I 3) have not found (not/find) one I like.

**My Spanish lessons are going very well. I 1) have been learning (learn) Spanish for five months now and I love it. 2) have already learned (already/learn) a lot.

**John 1) has been (be) very busy recently. He 2) has been painting (paint) the living room and the bedrooms, but he 3) has not started (not/start) painting the kitchen yet.

A

**I'm planning to buy a new house. I 1) have been looking (look) for one for two months now. So far I 2) have looked (look) at ten houses, but I 3) have not found (not/find) one I like.

B

**My Spanish lessons are going very well. I 1) have been learning (learn) Spanish for five months now and I love it. 2) have already learned (already/learn) a lot.

C

**John 1) has been (be) very busy recently. He 2) has been painting (paint) the living room and the bedrooms, but he 3) has not started (not/start) painting the kitchen yet.

ĐK: `x>=0` 

Ta có:

\(B=\dfrac{5\sqrt{x}-1}{\sqrt{x}+1}\\ =\dfrac{\left(5\sqrt{x}+5\right)-6}{\sqrt{x}+1}=\dfrac{5\left(\sqrt{x}+1\right)-6}{\sqrt{x}+1}\\ =\dfrac{5\left(\sqrt{x}+1\right)}{\sqrt{x}+1}-\dfrac{6}{\sqrt{x}+1}\\ =5-\dfrac{6}{\sqrt{x}+1}\)

Vì: \(\sqrt{x}\ge0\forall x\)

\(=>\sqrt{x}+1\ge1\forall x=>\dfrac{6}{\sqrt{x}+1}\le6\\ =>5-\dfrac{6}{\sqrt{x}+1}\ge5-6=-1\)

Dấu "=" xảy ra: `x=0` 

Ta có pt hoành độ giao điểm là: 

\(-x^2=\left(2-m\right)x+m-3\\ \Leftrightarrow x^2+\left(2-m\right)x+m-3=0\)

Để pt có nghiệm phân biệt thì: 

\(\Delta=\left(2-m\right)^2-4\cdot1\cdot\left(m-3\right)\\ =4-4m+m^2-4m+12=m^2-8m+16=\left(m-4\right)^2>0\) 

`=>m-4<>0<=>m<>4` 

Ta có: `a+b+c=1+(2-m)+(m-3)=0`

\(=>x_1=1\)
Theo vi-ét ta có: \(x_1+x_2=m-2=>x_2=m-2-x_2=m-2-1=m-3\) 

\(\left|x_1\right|+x_2^2=2\\ =>1+\left(m-3\right)^2=2\\< =>\left(m-3\right)^2=2-1=1\\ < =>\left[{}\begin{matrix}m-3=1\\m-3=-1\end{matrix}\right.\\ < =>\left[{}\begin{matrix}m=1+3=4\left(ktm\right)\\m=-1+3=2\left(tm\right)\end{matrix}\right.\)

Vậy: ... 

ΔAHB vuông tại H

=>\(HA^2+HB^2=AB^2\)

=>\(HB=\sqrt{15^2-9^2}=12\left(cm\right)\)

ΔABC cân tại A

mà AH là đường cao

nên H là trung điểm của BC

=>\(BC=2\cdot BH=24\left(cm\right)\)

Xét ΔABC có \(cosBAC=\dfrac{AB^2+AC^2-BC^2}{2\cdot AB\cdot AC}=\dfrac{15^2+15^2-24^2}{2\cdot15\cdot15}=\dfrac{-7}{25}\)

=>\(sinBAC=\sqrt{1-\left(-\dfrac{7}{25}\right)^2}=\sqrt{1-\dfrac{49}{625}}=\dfrac{24}{25}\)

Xét ΔABC có \(\dfrac{BC}{sinBAC}=2R\)

=>\(2R=24:\dfrac{24}{25}=25\)

=>R=12,5(cm)

A = (\(\dfrac{1}{x-\sqrt{x}}\)  + \(\dfrac{1}{\sqrt{x}+1}\)) : \(\sqrt{x}\) + \(\dfrac{1}{x-2\sqrt{x}+1}\)

Có phải đề bài như này không em?

a, PT: \(Zn+CuSO_4\rightarrow ZnSO_4+Cu\)

_______x_______________x_______x (mol)

Ta có: m _giảm = m_Zn - m_Cu

⇒ 20 - 19,96 = 65x - 64x

⇒ x = 0,04 (mol)

⇒ m_{Zn (pư)} = 0,04.65 = 2,6 (g)

b, m_ZnSO4 = 0,04.161 = 6,44 (g)

a.

Gọi x là số mol kẽm tham gia phản ứng.

\(Zn+CuSO_4\rightarrow ZnSO_4+Cu\)

 x ---->x----------->x-------->x

Ta có:

\(m_{Zn}-m_{Cu}=m_{kl.giảm}\Leftrightarrow65x-64x=20-19,96\Leftrightarrow x=0,04\left(mol\right)\)

=> \(m_{Zn.pứ}=0,04.65=2,6\left(g\right)\)

b)

\(m_{ZnSO_4}=0,04.160=6,4\left(g\right)\)