Bài 1.

\(a\Big) 9(4x+3)^2=16(3x-5)^2\\\Leftrightarrow 9[(4x)^2+2\cdot 4x\cdot3+3^2]=16[(3x)^2-2\cdot3x\cdot5+5^2]\\\Leftrightarrow9(16x^2+24x+9)=16(9x^2-30x+25)\\\Leftrightarrow 144x^2+216x+81=144x^2-480x+400\\\Leftrightarrow (144x^2-144x^2)+(216x+480x)=400-81\\\Leftrightarrow 696x=319\\\Leftrightarrow x=\dfrac{11}{24}\\Vậy:x=\dfrac{11}{24}\\---\)

\(b\Big)(x-3)^2=4x^2-20x+25\\\Leftrightarrow(x-3)^2=(2x)^2-2\cdot2x\cdot5+5^2\\\Leftrightarrow(x-3)^2=(2x-5)^2\\\Leftrightarrow (x-3)^2-(2x-5)^2=0\\\Leftrightarrow (x-3-2x+5)(x-3+2x-5)=0\\\Leftrightarrow (-x+2)(3x-8)=0\\\Leftrightarrow \left[\begin{array}{} -x+2=0\\ 3x-8=0 \end{array} \right.\\\Leftrightarrow \left[\begin{array}{} -x=-2\\ 3x=8 \end{array} \right.\\\Leftrightarrow \left[\begin{array}{} x=2\\ x=\dfrac{8}{3} \end{array} \right.\\Vậy:...\)

\(a,\Rightarrow4x^2+20x+25=0\Rightarrow\left(2x+5\right)^2=0\Rightarrow2x+5=0\Rightarrow x=-\dfrac{5}{2}\\ b,\Rightarrow x^3-6x^2+12x-8=0\Rightarrow\left(x-2\right)^3=0\Rightarrow x-2=0\Rightarrow x=2\)

a) \(\Rightarrow4x^2+20x+25=0\)

\(\Rightarrow\left(2x+5\right)^2=0\Rightarrow2x+5=0\)

\(\Rightarrow x=-\dfrac{5}{2}\)

b) \(\Rightarrow x^3-6x^2+12x-8=0\)

\(\Rightarrow\left(x-2\right)^3=0\Rightarrow x=2\)

Bài 2: 

a: =>4x(x+5)=0

=>x=0 hoặc x=-5

b: =>(x+3)(x-3)=0

=>x=-3 hoặc x=3

a,\(Đkxđ:x\ge3\)

Ta có:

\(\sqrt{\left(x-3\right)^2}=3-x\)

\(\Leftrightarrow|x-3|=3-x\)

\(\Leftrightarrow x-3=\left[{}\begin{matrix}x-3\\3-x\end{matrix}\right.\)

\(TH1:x-3=x-3\Leftrightarrow0x=0\)

\(\Rightarrow\)\(x\in R\) và \(x\ge3\)

\(TH2:x-3=3-x\Leftrightarrow2x=6\Leftrightarrow x=3\)( ko thỏa mãn điều kiện)

vậy \(\left\{x\in R/x\ge3\right\}\)

b, \(Đkxđ:x\le\dfrac{5}{2}\)

Ta có:

\(\sqrt{25-20x+4x^2}+2x=5\)

\(\Leftrightarrow\sqrt{\left(5-2x\right)^2}+2x=5\)

\(\Leftrightarrow\left|5-2x\right|=5-2x\)

\(\Leftrightarrow\left[{}\begin{matrix}5-2x=5-2x\\5-2x=2x-5\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}0x=0\\4x=10\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x\in R\\x=\dfrac{5}{2}\left(tmđk\right)\end{matrix}\right.\)

Vậy \(\left\{x\in R/x\le\dfrac{5}{2}\right\}\)

Chúc bạn học tốt!

để đâu bn

DƯƠNG PHAN KHÁNH DƯƠNG

Phùng Khánh Linh

Aki Tsuki

@Ngô Thanh Sang

\(\sqrt{25-20x+4x^2}+2x=5\)

\(\Leftrightarrow\sqrt{25-20x+4x^2}=5-2x\) ( ĐK : \(x\ge\dfrac{5}{2}\) )

\(\Leftrightarrow\sqrt{\left(2x-5\right)^2}=5-2x\)

\(\Leftrightarrow\left|2x-5\right|=5-2x\)

Với \(x\ge\dfrac{5}{2}\) :

\(\Leftrightarrow2x-5=5-2x\)

\(\Leftrightarrow4x=10\)

\(\Leftrightarrow x=\dfrac{5}{2}\) ( Thỏa mãn )

Với \(x< \dfrac{5}{2}\) :

\(\Leftrightarrow-2x+5=5-2x\)

\(\Leftrightarrow0=0\)

Vậy \(x=\dfrac{5}{2}\)

a)

\(\sqrt{3x^2+6x+7}+\sqrt{5x^2+10x+21}=5-2x-x^2\)

\(\Leftrightarrow\sqrt{3\left(x+1\right)^2+4}+\sqrt{5\left(x+1\right)^2+16}=6-\left(x+1\right)^2\)

\(VT\ge6;VP\le6\Rightarrow VT=VP=6\)

Vậy pt có một nghiệm duy nhất là \(x=-1\)

b)

\(\sqrt{4x^2+20x+25}+\sqrt{x^2-8x+16}=\sqrt{x^2+18x+81}\)

\(\Leftrightarrow\sqrt{\left(2x+5\right)^2}+\sqrt{\left(x-4\right)^2}=\sqrt{\left(x+9\right)^2}\)

\(\Leftrightarrow\left|2x+5\right|+\left|x-4\right|=\left|x+9\right|\)

Lập bảng xét dấu ra nhé ~^o^~