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Đổi 18 lít = 0,018 m³

Trọng lượng của 18 lít dầu hoả là :

\(0,018.\left(800.10\right)=144\left(N\right)\)

Vậy .........

TẤT CẢ CÁC CÂU TRÊN ĐỀU ĐÚNG

Dổi 1lít = 0,001 m³

3 lít nước sẽ có trọng lượng là :

áp dụng công thức :

\(d=\frac{P}{V}\)ta có

\(d=\frac{1000.10}{0,001.3}=\frac{10000}{0,003}\)( N )

vậy....

1kg sắt sẽ có thể tích là :

\(1:7800=\frac{1}{7800}\left(m^3\right)\)

vậy.............

Reported speech

1. " Hơ many jobs have you had since you left school?" the interviewer asked Mr Simpson.

-> The interviewer ask Mr Simpson how many jobs he had had since he left school.

2. " why didn't you report the incident to the police last week?" the officer asked the frightened witness.

-> The officer wanted to know why hadn't the frightened witness reported the incident to the police the week before.

3. "Where are you spendring your holiday?" Janet asked us.

-> Janet asked us where we were spendring our holiday.

1. " How many jobs have you had since you left school?" the interviewer asked Mr Simpson.

-> The interviewer asked ..Mr Simpson how many jobs he/she had since he/she left school...............................

2. " why didn't you report the incident to the police last week?" the officer asked the frightened witness.

-> The officer wanted to know ....why frightened witness had reported the icident to the police the week before.........

3. "Where are you spendring your holiday?" Janet asked us.

-> Janet asked ...us where we were spending our holiday...........

=)) có gì sai sót mong thí chủ thông cảm, lâu k làm cái này

COMPLETE THE SENTENCES WITH THE CORRECT FORM OF THE ADJ / ADV IN BRACKETSM

1) He goes to the theater ( more often ) ....... than i do .

2) They work ( harder ) ....... than us .

1, often

2, harder

Lời giải:

a) Sử dụng tính chất đường phân giác (đường phân giác $BD, AI$) ta có:

\(\bullet \frac{AD}{DC}=\frac{AB}{BC}\Rightarrow \frac{AD}{AD+DC}=\frac{AD}{AC}=\frac{AB}{BC+AB}\)

\(\Rightarrow AD=\frac{AB.AC}{BC+AB}(1)\)

\(\bullet \frac{BI}{ID}=\frac{AB}{AD}\Rightarrow \frac{BI}{ID+BI}=\frac{BI}{BD}=\frac{AB}{AD+AB}(2)\)

Từ \((1);(2)\Rightarrow \frac{BI}{BD}=\frac{AB}{\frac{AB.AC}{BC+AB}+AB}=\frac{BC+AB}{AC+BC+AB}\) (đpcm)

b)

\(BI.IC=\frac{1}{2}BD.CI\Leftrightarrow \frac{BI}{BD}=\frac{1}{2}\)

\(\Leftrightarrow \frac{AB+BC}{AB+BC+AC}=\frac{1}{2}\)

\(\Leftrightarrow AC=AB+BC\) (trái với BĐT tam giác ) nên bạn xem lại đề.

Lời giải:

Áp dụng BĐT Bunhiacopxky:

\(P^2=(a\sqrt{15ab+10b^2}+b\sqrt{15ab+10a^2})^2\leq (a^2+b^2)(15ab+10b^2+15ab+10a^2)\)

\(P^2\leq (a^2+b^2)(30ab+10a^2+10b^2)\)

Áp dụng BĐT Cauchy: \(2ab\leq a^2+b^2\Rightarrow 30ab\leq 15(a^2+b^2)\)

Do đó: \(P^2\leq (a^2+b^2)(15a^2+15b^2+10a^2+10b^2)=25(a^2+b^2)^2\)

\(\Rightarrow P\leq 5(a^2+b^2)\leq 5.2=10\)

Vậy $P_{\max}=10$ khi $a=b=1$

a³ + b³ + c³ = 3abc

<=> a³ + b³ + c³ - 3abc = 0

<=> (a + b + c)(a² + b² + c² - ab - bc - ca) = 0

<=> (a + b + c)[(a - b)² + (b - c)² + (c - a)²] = 0

<=> \(\left[{}\begin{matrix}a+b+c=0\Rightarrow\left\{{}\begin{matrix}a+b=-c\\b+c=-a\\c+a=-b\end{matrix}\right.\\\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\Leftrightarrow a=b=c\end{matrix}\right.\)

TH1: a + b + c = 0

\(A=\left(1+\dfrac{a}{b}\right)\left(1+\dfrac{b}{c}\right)\left(1+\dfrac{c}{a}\right)=\dfrac{a+b}{b}.\dfrac{b+c}{c}.\dfrac{c+a}{a}=\dfrac{-c}{b}.\dfrac{-a}{c}.\dfrac{-b}{a}=-1\)

TH2: a = b = c

A = 2.2.2 = 8

“Will you come to my party?” she said to me.

=>She asked me..if /whether I would come to her party.

“Will you come to my party?” she said to me.

=>She invited me to come to her party

I. Điền giới từ thích hợp :

1. The thief, Peter Russel, has escaped.......from......... Longbay Prison. L

2.You shouldn't fight..........between......your brothers and sisters.

3. The lilifeguard saved the woman......from............drowning.

4. The search.......of..........the lost boy lasted thirty six hours.

5. You shouldn't have lied......to.......your friends....about........your exam results.

6. Can you wait.......for.......me? I won't be long.

II. Rewrite :

1. Linh likes the country life more than the city life.(prefers)

-> Linh.....prefers the country life to the city life....................................

2. He would go hunting when he was in Africa.

-> He used.....to go hunting when he was in Africa..............................

III. Chuyển sang dạng bị động.

- You can make payment at any post office.

->Payment......can be made at any post office...........................

His dog is called Tony.

-> He.............................

Còn câu này nữa các bạn giúp mik vs

Do \(a,b,c\) là các số dương suy ra:

\(a>0;b>0;c>0\)

Suy ra: \(a+b+c>0\)

Ta có: \(a^3+b^3+c^3=3abc\)

\(\Leftrightarrow\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc=0\)

\(\Leftrightarrow\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2-3ab\left(a+b+c\right)\right]=0\)

\(\Leftrightarrow\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)=0\)

\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=0\)

\(\Rightarrow a+b+c=0\) hoặc \(a^2+b^2+c^2-ab-bc-ca=0\)

Do \(a+b+c>0\)

Suy ra: \(a^2+b^2+c^2-ab-bc-ca=0\)

\(2a^2+2b^2+2c^2-2ab-2bc-2ca=0\)

\(\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ca+a^2\right)=0\)

\(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)

Suy ra: \(a-b=0;b-c=0\) và \(c-a=0\)

Suy ra: \(a=b=c\)

Suy ra: \(\frac{a}{b}=\frac{b}{c}=\frac{c}{a}=1\)

Ta có: \(\left(\frac{a}{b}-1\right)+\left(\frac{b}{c}-1\right)+\left(\frac{c}{a}-1\right)=\left(1-1\right)+\left(1-1\right)+\left(1-1\right)=0\)

Vậy ...

Lời giải:

ĐKXĐ: \(x\geq \frac{6}{5}\)

PT \(\Leftrightarrow 2(x+1)\sqrt{5x-6}=(x+1)^2+5x-6-1\)

\(\Leftrightarrow (x+1)^2+(5x-6)-2(x+1)\sqrt{5x-6}-1=0\)

\(\Leftrightarrow (x+1-\sqrt{5x-6})^2-1=0\)

\(\Leftrightarrow (x+2-\sqrt{5x-6})(x-\sqrt{5x-6})=0\)

\(\Rightarrow \left[\begin{matrix} x+2=\sqrt{5x-6}\\ x=\sqrt{5x-6}\end{matrix}\right.\)

Nếu \(x+2=\sqrt{5x-6}\Rightarrow \left\{\begin{matrix} x\geq -2\\ (x+2)^2=5x-6\end{matrix}\right.\)

\(\Leftrightarrow \left\{\begin{matrix} x\geq -2\\ x^2-x+10=0\end{matrix}\right.\) (dễ thấy vô nghiệm)

Nếu \(x=\sqrt{5x-6}\Rightarrow \left\{\begin{matrix} x\geq 0\\ x^2=5x-6\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq 0\\ x^2-5x+6=0\end{matrix}\right.\)

\(\Leftrightarrow \left\{\begin{matrix} x\geq 0\\ (x-2)(x-3)=0\end{matrix}\right.\Rightarrow x=2; x=3\) là nghiệm của PT

Vậy.......

Lời giải:

Áp dụng BĐT Cauchy cho các số không âm:

\((x-1)+4\geq 2\sqrt{4(x-1)}=4\sqrt{x-1}\)

\(\Leftrightarrow \frac{x+3}{4}\geq \sqrt{x-1}(1)\)

\((y+1)+\frac{1}{4}\geq 2\sqrt{(y+1).\frac{1}{4}}=\sqrt{y+1}(2)\)

Lấy \((1)+(2)\Rightarrow \frac{x}{4}+y+2\geq \sqrt{x-1}+\sqrt{y+1}\)

Dấu "=" xảy ra (như trong điều kiện đề bài) khi :

\(\left\{\begin{matrix} x-1=4\\ y+1=\frac{1}{4}\end{matrix}\right.\Leftrightarrow x=5; y=-\frac{3}{4}\)