bằng phuiong pháp đặt ẩn phụ hoàn toàn nha

Đặt \(x^2+3x=t\left(t>0\right)\)

Ta có: \(\sqrt{x^2+3x+12}=x^2+3x\)

\(\Rightarrow\sqrt{t+12}=t\)

\(\Rightarrow t+12=t^2\)

\(\Rightarrow t^2-t-12=0\)

\(\Rightarrow t\left(t-4\right)+3\left(t-4\right)=0\)

\(\Rightarrow\left(t-4\right)\left(t+3\right)=0\Rightarrow\orbr{\begin{cases}t=4\left(t/m\right)\\t=-3\left(loai\right)\end{cases}}\)

Ta có: \(x^2+3x=4\)

\(\Rightarrow x^2+3x-4=0\)

\(\Rightarrow x\left(x+4\right)-\left(x+4\right)=0\)

\(\Rightarrow\left(x+4\right)\left(x-1\right)=0\Rightarrow\orbr{\begin{cases}x=-4\\x=1\end{cases}}\)

Chúc bạn học tốt.

Chia đa thức ta được số dư là a + 8

\(\Rightarrow a+8=0\Rightarrow a=-8\)

8 thôi chứ bạn

bạn giúp mk bài văn của mk nha bn

Bài 1

a) \(3x\left(4x^2-2x+3\right)\)

\(=3x.4x^2-3x.2x+3x.3\)

\(=12x^3-6x^2+9x\)

b) \(\left(2x+5\right)^2-4x^2\)

\(=\left[\left(2x+5\right)-4x\right]\left[\left(2x+5\right)+4x\right]\)

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

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

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

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

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

Bài 2

a) \(6x^2y+18x\)

\(=6x\left(xy+3\right)\)

b) \(x^2-7x+3x-21\)

\(=\left(x^2-7x\right)+\left(3x-21\right)\)

\(=x\left(x-7\right)+3\left(x-7\right)\)

\(=\left(x-7\right)\left(x+3\right)\)

c) \(x^2-4y^2+2x+1\)

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

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

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

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

\(=\left[\left(x+1\right)-2y\right]\left[\left(x+1\right)+2y\right]\)

\(=\left(x+1-2y\right)\left(x+1+2y\right)\)

d) \(x^2+3x-3y-y^2\)

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

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

\(=\left(x-y\right)\left[\left(x+y\right)+3\right]\)

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

Bài 3

a) \(\left(x+3\right)\left(x+2\right)-x\left(x+3\right)=10\)

\(\Rightarrow\left(x+3\right)\left[\left(x+2\right)-x\right]=10\)

\(\Rightarrow\left(x+3\right)\left(x+2-x\right)=10\)

\(\Rightarrow\left(x+3\right).2=10\)

\(\Rightarrow x+3=5\)

\(\Rightarrow x=2\)

b) \(\left(x+2\right)^2-\left(x-3\right)\left(x+3\right)=10\)

\(\Rightarrow\left(x^2+2.x.2+2^2\right)-\left(x^2-3^2\right)=10\)

\(\Rightarrow\left(x^2+4x+4\right)-\left(x^2-9\right)=10\)

\(\Rightarrow x^2+4x+4-x^2+9=10\)

\(\Rightarrow4x+13=10\)

\(\Rightarrow4x=-3\)

\(\Rightarrow x=-\frac{3}{4}\)

c) \(4x^2-25=0\)

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

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

\(\Rightarrow2x-5=0\) hoặc \(2x+5=0\)

\(\Rightarrow2x=5\)          hoặc\(2x=-5\)

\(\Rightarrow x=\frac{5}{2}\)          hoặc\(x=-\frac{5}{2}\)

d) \(2x\left(x+3\right)+x^2+3x=0\)

\(\Rightarrow2x\left(x+3\right)+x\left(x+3\right)=0\)

\(\Rightarrow\left(x+3\right)\left(2x+x\right)=0\)

\(\Rightarrow\left(x+3\right).3x=0\)

\(\Rightarrow x+3=0\)  hoặc \(3x=0\)

\(\Rightarrow x=-3\)      hoặc \(x=0\)

K MÌNH VỚI NHÉ 

💕💕💕Thanks bn nhìu nhìu nha😻😻😻😻

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