1: =>(x-4)(x+1)=0

=>x=4 hoặc x=-1

2: =>x²-5x+6=0

=>(x-2)(x-3)=0

=>x=2 hoặc x=3

3: =>7x²+7x-x-1=0

=>(x+1)(7x-1)=0

=>x=-1 hoặc x=1/7

\(a,\sqrt{x}< 5\Leftrightarrow x< 25\\ b,\sqrt{x}=10\Leftrightarrow x=100\\ c,\sqrt{x^2}=7\Leftrightarrow\left|x\right|=7\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-7\end{matrix}\right.\\ d,\sqrt{x^2}=\left|-8\right|\Leftrightarrow\left|x\right|=8\Leftrightarrow\left[{}\begin{matrix}x=8\\x=-8\end{matrix}\right.\)

a) \(\sqrt{x}< 5\text{⇒}x< 25\)

b) \(\sqrt{x}=10\text{⇒}x=100\)

c) \(\sqrt{x^2}=7\text{⇒}x^2=49\text{⇒}x=+-7\)

d) \(\sqrt{x^2}=\left|-8\right|\text{⇒}x^2=64\text{⇒}x=+-8\)

\(\sqrt{2x^2-3x-5}=\sqrt{x^2-7}\)

Bình phương 2 vế pt , ta có :

\(2x^2-3x-5=x^2-7\)

\(\Rightarrow2x^2-x^2-3x=-7+5\)

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

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3+\sqrt{17}}{2}\\x=\dfrac{3-\sqrt{17}}{2}\end{matrix}\right.\)

Thay lần lượt các giá trị trên vào pt , ta thấy không có giá trị nào thỏa mãn

Vậy pt vô nghiệm

=>2x^2-3x-5=x^2-7

=>x^2-3x+2=0

=>x=2(loại) hoặc x=1(loại)

a: Khi m=2 thì (1) sẽ là x^2+2x+1=0

=>x=-1

b:x1+x2=52

=>2m-2=52

=>2m=54

=>m=27

\(P-8=\dfrac{x^2-6x+9}{x-1}=\dfrac{\left(x-3\right)^2}{x-1}\ge0\) (Do x > 1 và \(\left(x-3\right)^2\ge0\forall x\in R\)).

Do đó \(P\ge8\). Dấu "=" xảy ra khi x = 3.

1) Ta có: \(3x\left(2-5x\right)+35x-14=0\)

\(\Leftrightarrow3x\left(2-5x\right)+7\left(5x-2\right)=0\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}5x-2=0\\-3x+7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}5x=2\\-3x=-7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{5}\\x=\dfrac{7}{3}\end{matrix}\right.\)

Vậy: \(S=\left\{\dfrac{2}{5};\dfrac{7}{3}\right\}\)

2) Ta có: \(4x-6+5x\left(3-2x\right)=0\)

\(\Leftrightarrow2\left(2x-3\right)-5x\left(2x-3\right)=0\)

\(\Leftrightarrow\left(2x-3\right)\left(2-5x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-3=0\\2-5x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=3\\5x=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{2}{5}\end{matrix}\right.\)

Vậy: \(S=\left\{\dfrac{3}{2};\dfrac{2}{5}\right\}\)

a: Ta có: \(\left\{{}\begin{matrix}3x+2y=14\\5x+3y=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}15x+10y=70\\15x+9y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=67\\3x=14-2y=14-2\cdot67=-120\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-40\\y=67\end{matrix}\right.\)

b: Ta có: \(\left\{{}\begin{matrix}-x+2y-6=0\\5x-3y-5=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}-x+2y=6\\5x-3y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-5x+10y=30\\5x-3y=5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}7y=35\\2y-x=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=5\\y=4\end{matrix}\right.\)

\(x^2-5x-14=0\)

\(\Leftrightarrow x^2+2x-7x-14=0\)

\(\Leftrightarrow x\left(x+2\right)-7\left(x+2\right)=0\)

\(\Leftrightarrow\left(x+2\right)\left(x-7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\x-7=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=7\end{matrix}\right.\)

Vậy: \(S=\left\{-2;7\right\}\)

\(x^2-5x-14=0\)

=>\(x^2-7x+2x-14=0\)

=>\(\left(x-7\right)\left(x+2\right)=0\)

=>\(\left[{}\begin{matrix}x-7=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-2\end{matrix}\right.\)

`x^2 -5x-14=0`

`<=>x^2 +2x-7x-14=0`

`<=> (x^2 +2x)-(7x+14)=0`

`<=> x (x+2)-7(x+2)=0`

`<=>(x+2)(x-7)=0`

\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\x-7=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-2\\x=7\end{matrix}\right.\)

lần sau bạn ghi rõ đề bài nhé

a, \(A=x^2+10x+39=x^2+2.5.x+25+14=\left(x+5\right)^2+14\ge14\)

Dấu ''='' xảy ra khi \(x=-5\)

Vậy GTNN A là 14 khi x = -5

b, \(B=25x^2-70x+1000=\left(5x\right)^2-2.5x.7+49+951\)

\(=\left(5x-7\right)^2+951\ge951\)

Dấu ''='' xảy ra khi \(x=\frac{7}{5}\)

Vậy GTNN B là 951 khi x = 7/5 

c, \(C=49x^2+64x+100=\left(7x\right)^2+2.7x.\frac{32}{7}+\frac{1024}{49}+\frac{3876}{49}\)

\(=\left(7x+\frac{32}{7}\right)^2+\frac{3876}{49}\ge\frac{3876}{49}\)

Dấu ''='' xảy ra khi \(x=-\frac{32}{49}\)

Vậy GTNN C là 3876/49 khi x = -32/49 

d, \(D=5x^2+13x+41=5\left(x^2+\frac{13}{5}x+\frac{41}{5}\right)\)

\(=5\left(x^2+2.\frac{6,5}{5}.x+\frac{169}{100}+\frac{651}{100}\right)=5\left(x+\frac{6,5}{5}\right)^2+\frac{651}{20}\ge\frac{651}{20}\)

Dấu ''='' xảy ra khi \(x=-\frac{6,5}{2}\)

Vậy GTNN D là 651/20 khi x = -6,5/2