Đăng từng bài thôi nha bạn 

Bài 1 : Năm nay mới lên lớp 8 -_- 

Bài 2 : 

\(a)\) 

* Câu A : 

\(A=x^2+4x-7\)

\(A=\left(x^2+4x+4\right)-11\)

\(A=\left(x+2\right)^2-11\ge-11\)

Dấu "=" xảy ra \(\Leftrightarrow\)\(x=-2\) ( ở đây nhiều bài quá nên mình làm tắt cho nhanh, bạn nhớ trình bày rõ ra nhé ) 

Vậy GTNN của \(A\) là \(-11\) khi \(x=-2\)

* Câu B : 

\(B=2x^2-3x+5\)

\(2B=4x^2-6x+10\)

\(2B=\left(4x^2-6x+1\right)+9\)

\(2B=\left(2x-1\right)^2+9\ge9\)

\(B=\frac{\left(2x-1\right)^2+9}{2}\ge\frac{9}{2}\)

Dấu "=" xảy ra \(\Leftrightarrow\)\(x=\frac{1}{2}\)

Vậy GTNN của \(B\) là \(\frac{9}{2}\) khi \(x=\frac{1}{2}\)

* Câu C : 

\(C=x^4-3x^2+1\)

\(C=\left(x^4-3x^2+\frac{9}{4}\right)-\frac{5}{4}\)

\(C=\left(x^2-\frac{3}{2}\right)^2-\frac{5}{4}\ge-\frac{5}{4}\)

Dấu "=" xảy ra \(\Leftrightarrow\)\(\orbr{\begin{cases}x=\sqrt{\frac{3}{2}}\\x=-\sqrt{\frac{3}{2}}\end{cases}}\)

Vậy GTNN của \(C\) là \(-\frac{5}{4}\) khi \(x=\sqrt{\frac{3}{2}}\) hoặc \(x=-\sqrt{\frac{3}{2}}\)

Chúc bạn học tốt ~ 

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5.\(C\text{ó}x^2-12=0\Rightarrow x^2=12\Rightarrow x=\sqrt{12}ho\text{ặc}x=-\sqrt{12}\)

Mà x>0\(\Rightarrow x=\sqrt{12}\)

6.Vì x-y=4\(\Rightarrow\left(x-y\right)^2=x^2-2xy+y^2=x^2-10+y^2=4^2=16\Rightarrow x^2+y^2=26\)

Có \(\left(x+y\right)^2=x^2+2xy+y^2=26+10=36=6^2=\left(-6\right)^2\)

Vì xy>0 và x>0 =>y>0=>x+y>0=>x+y=6

7. \(3x^2+7=\left(x+2\right)\left(3x+1\right)\)

\(3x^2+7=3x^2+7x+2\)

\(3x^2+7-3x^2-7x-2=0\)

-7x+5=0

-7x=-5

\(x=\frac{5}{7}\)

8.\(\left(2x+1\right)^2-4\left(x+2\right)^2=9\)

\(\left(2x+1\right)^2-\left(2x+4\right)^2=9\)

(2x+1-2x-4)(2x+1+2x+4)=9

-3(4x+5)=9

4x+5=-3

4x=-8

x=-2

Còn câu 9 và 10 để mình nghiên cứu đã

biet x+y =2 tinh min 3x^2 + y^2

tích cho tớ nha cậu, mơn nhìu ạk

Ai biết cách làm thì nhanh tay giải giùm mình nhé!!!!!!!!!!!!

mk đang cần gấp....<3<3<3<3<3<3

) \(\dfrac{x^3+8y^3}{2y+x}\)

\(=\dfrac{x^3+\left(2y\right)^3}{x+2y}\)

\(=\dfrac{\left(x+2y\right)\left[x^2+x.2y+\left(2y\right)^2\right]}{x+2y}\)

\(=x^2+2xy+4y^2\)

b) \(\dfrac{a-1}{2\left(a-4\right)}+\dfrac{a}{a-4}\) MTC: \(2\left(a-4\right)\)

\(=\dfrac{a-1}{2\left(a-4\right)}+\dfrac{2a}{2\left(a-4\right)}\)

\(=\dfrac{a-1+2a}{2\left(a-4\right)}\)

\(=\dfrac{3a-1}{2\left(a-4\right)}\)

c) \(\dfrac{x^3+3x^2y+3xy^2+y^3}{2x+2y}\)

\(=\dfrac{\left(x+y\right)^3}{2\left(x+y\right)}\)

\(=\dfrac{\left(x+y\right)^2}{2}\)

d) \(\left(x-5\right)^2+\left(7-x\right)\left(x+2\right)\)

\(=\left(x^2-2.x.5+5^2\right)+\left(7x+14-x^2-2x\right)\)

\(=x^2-10x+25+7x+14-x^2-2x\)

\(=39-5x\)

e) \(\dfrac{3x}{x-2}-\dfrac{2x+1}{2-x}\)

\(=\dfrac{3x}{x-2}+\dfrac{2x+1}{x-2}\)

\(=\dfrac{3x+2x+1}{x-2}\)

\(=\dfrac{5x+1}{x-2}\)

h) \(\dfrac{1}{3x-2}-\dfrac{1}{3x+2}-\dfrac{3x+6}{4-9x^2}\)

\(=\dfrac{1}{3x-2}-\dfrac{1}{3x+2}+\dfrac{3x+6}{9x^2-4}\)

\(=\dfrac{1}{3x-2}-\dfrac{1}{3x+2}+\dfrac{3x+6}{\left(3x-2\right)\left(3x+2\right)}\) MTC: \(\left(3x-2\right)\left(3x+2\right)\)

\(=\dfrac{3x+2}{\left(3x-2\right)\left(3x+2\right)}-\dfrac{3x-2}{\left(3x-2\right)\left(3x+2\right)}+\dfrac{3x+6}{\left(3x-2\right)\left(3x+2\right)}\)

\(=\dfrac{\left(3x+2\right)-\left(3x-2\right)+\left(3x+6\right)}{\left(3x-2\right)\left(3x+2\right)}\)

\(=\dfrac{3x+2-3x+2+3x+6}{\left(3x-2\right)\left(3x+2\right)}\)

\(=\dfrac{3x+10}{\left(3x-2\right)\left(3x+2\right)}\)

câu f ,g đâu

b) Ta có: \(\left(x-2\right)^3+\left(3x-1\right)\left(3x+1\right)=\left(x+1\right)^3\)

⇔\(\left(x-2\right)^3+\left(3x-1\right)\left(3x+1\right)-\left(x+1\right)^3=0\)

⇔\(x^3-6x^2+12x-8+9x^2-1-\left(x^3+3x^2+3x+1\right)=0\)

⇔\(x^3+3x^2+12x-9-x^3-3x^2-3x-1=0\)

⇔\(9x-10=0\)

hay 9x=10

⇔\(x=\frac{10}{9}\)

Vậy: \(x=\frac{10}{9}\)

c) \(\frac{2x-1}{5}-\frac{x-2}{3}=\frac{x+7}{5}\)

⇔\(\frac{2x-1}{5}-\frac{x-2}{3}-\frac{x+7}{5}=0\)

⇔\(\frac{3\left(2x-1\right)}{15}-\frac{5\left(x-2\right)}{15}-\frac{3\left(x+7\right)}{15}=0\)

⇔\(3\left(2x-1\right)-5\left(x-2\right)-3\left(x+7\right)=0\)

⇔\(6x-3-5x+10-3x-21=0\)

⇔\(-2x-14=0\)

⇔\(-2x=14\)

hay x=-7

Vậy: x=-7

d) \(\frac{2\left(x-3\right)}{7}+\frac{x-5}{3}=\frac{13x+4}{21}\)

⇔\(\frac{2\left(x-3\right)}{7}+\frac{x-5}{3}-\frac{13x+4}{21}=0\)

⇔\(\frac{6\left(x-3\right)}{21}+\frac{7\left(x-5\right)}{21}-\frac{13x+4}{21}=0\)

⇔\(6x-18+7x-35-13x-4=0\)

⇔\(-21\ne0\)

Vậy: x∈∅

e) \(\frac{\left(x+10\right)\left(x+4\right)}{12}-\frac{\left(x+4\right)\left(2-x\right)}{4}=\frac{\left(x+10\right)\left(x-2\right)}{3}\)

⇔\(\frac{\left(x+10\right)\left(x+4\right)}{12}-\frac{\left(x+4\right)\left(2-x\right)}{4}-\frac{\left(x+10\right)\left(x-2\right)}{3}=0\)

⇔\(\frac{\left(x+10\right)\left(x+4\right)}{12}-\frac{3\left(x+4\right)\left(2-x\right)}{12}-\frac{4\left(x+10\right)\left(x-2\right)}{12}=0\)

⇔\(x^2+14x+40-\left(3x+12\right)\left(2-x\right)-\left(4x+40\right)\left(x-2\right)=0\)

⇔\(x^2+14x+40-\left(24-6x-3x^2\right)-\left(4x^2+32x-80\right)=0\)

⇔\(x^2+14x+40-24+6x+3x^2-4x^2-32x+80=0\)

⇔\(-12x+96=0\)

⇔\(-12x=-96\)

hay x=8

Vậy: x=8