MK KO BT MK MỚI HO C LỚP 6

AI HỌC LỚP 6 CHO MK XIN

Bài 3 : 

\(a)\left|3x-2\right|=x\)

\(\Rightarrow\orbr{\begin{cases}3x-2=x\\3x-2=-x\end{cases}\Rightarrow\orbr{\begin{cases}3x-x=2\\3x+x=2\end{cases}\Rightarrow}\orbr{\begin{cases}2x=2\\4x=2\end{cases}\Rightarrow}\orbr{\begin{cases}x=1\\x=\frac{1}{2}\end{cases}}}\)

vậy \(x=1;x=\frac{1}{2}\)

Bài 10

\(a)\)cách 1: cm vế trái bằng vế phải

\(\left(a-b\right)^2=\left(a-b\right)\left(a-b\right)\)

                  \(=a^2-ab-ab+b^2\)

                  \(=a^2-2ab+b^2\)

cách 2 : cm vế phải = vế trái

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

\(b)A=\left(5x^4-3y^3\right)^2\)

       \(=\left(5x^4\right)^2-2\times5x^4\times3y^3+\left(3y^3\right)^2\)

       \(=25x^8-30x^4y^3+9y^6\)

3.a.

ta có

 \(|3x-2|=x\\\Rightarrow\orbr{\begin{cases}3x-2=x\\-3x+2=x\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}3x-x=2\\-3x-x=-2\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}2x=2\\-4x=-2\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=1\\x=\frac{1}{2}\end{cases}}\)

10a:

ta có

\(\left(a-b\right)^2=\left(a-b\right)\left(a-b\right)\)

rồi nhân ra là dc

10b:

ta có 

\(\left(5x4-3y3\right)^2\)

\(=\left(20x-9y\right)^2\)

\(=\left(400x^2-2.20x.9y+81y^2\right)\)

rồi rút gọn là dc bạn ạ

a)Có \(a^2+1\ge2a\) với mọi a; \(b^2+1\ge2b\) với mọi b

Cộng vế với vế \(\Rightarrow a^2+b^2+2\ge2\left(a+b\right)\)

Dấu = xảy ra <=> a=b=1

b) Áp dụng BĐT bunhiacopxki có:

\(\left(x+y\right)^2\le\left(1+1\right)\left(x^2+y^2\right)\Leftrightarrow\left(x+y\right)^2\le2\)

\(\Leftrightarrow-\sqrt{2}\le x+y\le\sqrt{2}\)

\(\Rightarrow\left(x+y\right)_{max}=\sqrt{2}\Leftrightarrow\left\{{}\begin{matrix}x+y=\sqrt{2}\\x=y\end{matrix}\right.\)\(\Leftrightarrow x=y=\dfrac{\sqrt{2}}{2}\)

\(\left(x+y\right)_{min}=-\sqrt{2}\Leftrightarrow\left\{{}\begin{matrix}x+y=-\sqrt{2}\\x=y\end{matrix}\right.\)\(\Leftrightarrow x=y=-\dfrac{\sqrt{2}}{2}\)

c) \(S=\dfrac{1}{ab}+\dfrac{1}{a^2+b^2}=\dfrac{1}{a^2+b^2}+\dfrac{1}{2ab}+\dfrac{1}{2ab}\)

Với x,y>0, ta có: \(\dfrac{1}{x}+\dfrac{1}{y}\ge\dfrac{4}{x+y}\) (1)

Thật vậy (1) \(\Leftrightarrow\dfrac{y+x}{xy}\ge\dfrac{4}{x+y}\Leftrightarrow\left(x+y\right)^2\ge4xy\)\(\Leftrightarrow\left(x-y\right)^2\ge0\) (lđ)

Áp dụng (1) vào S ta được:

\(S\ge\dfrac{4}{a^2+b^2+2ab}+\dfrac{1}{2ab}\)

Lại có: \(ab\le\dfrac{\left(a+b\right)^2}{4}\) \(\Leftrightarrow2ab\le\dfrac{\left(a+b\right)^2}{2}\Leftrightarrow2ab\le\dfrac{1}{2}\)\(\Rightarrow\dfrac{1}{2ab}\ge2\)

\(\Rightarrow S\ge\dfrac{4}{\left(a+b\right)^2}+2=6\)

\(\Rightarrow S_{min}=6\Leftrightarrow a=b=\dfrac{1}{2}\)

a) f(x) = x(x - 5) + 2(x - 5)

x(x - 5) + 2(x - 5) = 0

<=> (x - 5)(x - 2) = 0

        x - 5 = 0 hoặc x - 2 = 0

        x = 0 + 5         x = 0 + 2

        x = 5               x = 2

=> x = 5 hoặc x = 2

a,   f(x) có nghiệm 

\(\Leftrightarrow x\left(x-5\right)+2\left(x-5\right)=0\)

\(\Rightarrow\left(x-5\right)\left(x+2\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-5=0\\x+2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=5\\x=-2\end{cases}}\)

->tự kết luận.

b1, để g(x) có nghiệm thì:

\(g\left(x\right)=2x\left(x-2\right)-x^2+5+4x=0\)

\(\Rightarrow2x^2-4x-x^2+5+4x=0\)

\(\Rightarrow x^2+5=0\)

Do \(x^2\ge0\forall x\)nên\(x^2+5\ge5\forall x\)

suy ra: k tồn tại \(x^2+5=0\)

Vậy:.....

b2, 

\(f\left(x\right)=x\left(x-5\right)+2\left(x-5\right)\)

\(=x^2-5x+2x-10\)

\(=x^2-3x-10\)

\(f\left(x\right)-g\left(x\right)=x^2+5-\left(x^2-3x-10\right)\)

\(=x^2+5-x^2+3x-10=3x-5\)

bài 5 nhé:

a) (a+1)²>=4a

<=>a²+2a+1>=4a

<=>a²-2a+1.>=0

<=>(a-1)²>=0 (luôn đúng)

vậy......

b) áp dụng bất dẳng thức cô si cho 2 số dương 1 và a ta có:

a+1>=\(2\sqrt{a}\)

tương tự ta có:

b+1>=\(2\sqrt{b}\)

c+1>=\(2\sqrt{c}\)

nhân vế với vế ta có:

(a+1)(b+1)(c+1)>=\(2\sqrt{a}.2\sqrt{b}.2\sqrt{c}\)

<=>(a+1)(b+1)(c+1)>=\(8\sqrt{abc}\)

<=>(a+)(b+1)(c+1)>=8 (vì abc=1)

vậy....

bạn nên viết ra từng câu

Chứ để như thế này khó nhìn lắm

bạn hỏi từ từ thôi

a: Ta có: \(x^2-36y^2-x+6y\)

\(=\left(x-6y\right)\left(x+6y\right)-\left(x-6y\right)\)

\(=\left(x-6y\right)\left(x+6y-1\right)\)

b: Ta có: \(16x-8x^2+x^3\)

\(=x\left(x^2-8x+16\right)\)

\(=x\left(x-4\right)^2\)

c: Ta có: \(2x^2-4xy+2y^2-18\)

\(=2\left(x^2-2xy+y^2-9\right)\)

\(=2\cdot\left[\left(x-y\right)^2-9\right]\)

\(=2\left(x-y-3\right)\left(x-y+3\right)\)

d: Ta có: \(3x^2-7x-10\)

\(=3x^2+3x-10x-10\)

\(=3x\left(x+1\right)-10\left(x+1\right)\)

\(=\left(x+1\right)\left(3x-10\right)\)

e: Ta có: \(x^4-x^2-30\)

\(=x^4-6x^2+5x^2-30\)

\(=x^2\left(x^2-6\right)+5\left(x^2-6\right)\)

\(=\left(x^2-6\right)\left(x^2+5\right)\)

f: Ta có: \(x^2-xy-2y^2\)

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

\(=x\left(x-2y\right)+y\left(x-2y\right)\)

\(=\left(x-2y\right)\left(x+y\right)\)

g: Ta có: \(x^4-13x^2y^2+4y^4\)

\(=x^4-4x^2y^2+4y^4-9x^2y^2\)

\(=\left(x^2-2y^2\right)^2-\left(3xy\right)^2\)

\(=\left(x^2-3xy-2y^2\right)\left(x^2-3xy+2y^2\right)\)

\(=\left(x^2-3xy-2y^2\right)\left(x^2-xy-2xy+2y^2\right)\)

\(=\left[x\left(x-y\right)-2y\left(x-y\right)\right]\left(x^2-3xy-2y^2\right)\)

\(=\left(x-y\right)\left(x-2y\right)\left(x^2-3xy-2y^2\right)\)

h: Ta có: \(\left(x^2-2x\right)^2-2\left(x^2-2x\right)-3\)

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

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

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

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

 a) 3x2 – 7x + 2

\(=3x^2-6x-x+2\)

\(=\left(3x^2-6x\right)-\left(x-2\right)\)

\(=3x\left(x-2\right)-\left(x-2\right)\)

\(=\left(x-2\right)\left(3x-1\right)\)

 b) a(x2 + 1) – x(a2 + 1)

\(=ax^2+a-\left(a^2x+x\right)\)

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

.......?

a) Ta có: \(3x^2-7x+2\)

\(=3x^2-6x-x+2\)

\(=3x\left(x-2\right)-\left(x-2\right)\)

\(=\left(x-2\right)\left(3x-1\right)\)

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

\(=x^2a+a-a^2x-x\)

\(=\left(x^2a-a^2x\right)+\left(a-x\right)\)

\(=xa\left(x-a\right)-\left(x-a\right)\)

\(=\left(x-a\right)\left(xa-1\right)\)

c) Ta có: \(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)

\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)

\(=\left(x^2+7x\right)^2+22\left(x^2+7x\right)+120-24\)

\(=\left(x^2+7x\right)^2+22\left(x^2+7x\right)+96\)

\(=\left(x^2+7x\right)^2+16\left(x^2+7x\right)+6\left(x^2+7x\right)+96\)

\(=\left(x^2+7x\right)\left(x^2+7x+16\right)+6\left(x^2+7x+16\right)\)

\(=\left(x^2+7x+16\right)\left(x^2+7x+6\right)\)

\(=\left(x^2+7x+16\right)\left(x+1\right)\left(x+6\right)\)

d) Ta có: \(\left(a+1\right)\left(a+3\right)\left(a+5\right)\left(a+7\right)+15\)

\(=\left(a^2+8a+7\right)\left(a^2+8a+15\right)+15\)

\(=\left(a^2+8a\right)^2+22\left(a^2+8a\right)+105+15\)

\(=\left(a^2+8a\right)^2+22\left(a^2+8a\right)+120\)

\(=\left(a^2+8a\right)^2+12\left(a^2+8a\right)+10\left(a^2+8a\right)+120\)

\(=\left(a^2+8a\right)\left(a^2+8a+12\right)+10\left(a^2+8a+12\right)\)

\(=\left(a^2+8a+12\right)\left(a^2+8a+10\right)\)

\(=\left(a+2\right)\left(a+6\right)\left(a^2+8a+10\right)\)