c: \(\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{7}\\x=-\sqrt{7}\\x=-5\\x=5\end{matrix}\right.\)

mik lớp 6 bạn

a: Đặt \(-10x^2+6x-7=0\)

=>\(10x^2-6x+7=0\)

\(\Delta=\left(-6\right)^2-4\cdot10\cdot7=36-40\cdot7=36-280=-244<0\)

mà a=-10<0

nên Biểu thức \(-10x^2+6x-7\) <0∀x

b: \(-4x^2+20x-25\)

\(=-\left(4x^2-20x+25\right)\)

\(=-\left\lbrack\left(2x\right)^2-2\cdot2x\cdot5+5^2\right\rbrack=-\left(2x-5\right)^2\le0\forall x\)

Dấu '=' xảy ra khi 2x-5=0

=>2x=5

=>x=5/2

c: \(-16x^2+56x-49\)

\(=-\left(16x^2-56x+49\right)\)

\(=-\left\lbrack\left(4x\right)^2-2\cdot4x\cdot7+7^2\right\rbrack=-\left(4x-7\right)^2\le0\forall x\)

Dấu '=' xảy ra khi 4x-7=0

=>4x=7

=>x=7/4

d: Đặt \(9x^2-1=0\)

=>\(9x^2=1\)

=>\(x^2=\frac19\)

=>x=1/3 hoặc x=-1/3

Đặt \(-2x^2+5x-2=0\)

=>\(2x^2-5x+2=0\)

=>\(2x^2-4x-x+2=0\)

=>(x-2)(2x-1)=0

=>x=2 hoặc x=1/2

Đặt \(F\left(x\right)=\left(9x^2-1\right)\left(-2x^2+5x-2\right)\)

Bảng xét dấu

b: -7<x<7

\(a.\sqrt{49-56x+16x^2}=\sqrt{16x^2-2.4x.7+49}=\sqrt{\left(4x-7\right)^2}=\text{|}4x-7\text{|}=4x-7\)\(b.\sqrt{x-2\sqrt{x-1}}=\sqrt{x-1-2\sqrt{x-1}+1}=\sqrt{\left(\sqrt{x-1}-1\right)^2}=\text{|}\sqrt{x-1}-1\text{|}=\sqrt{x-1}-1\)\(c.\sqrt{x+4\sqrt{x-4}}+\sqrt{x-4\sqrt{x-4}}=\sqrt{x-4+2.2\sqrt{x-4}+4}+\sqrt{x-4-2.2\sqrt{x-4}+4}=\sqrt{\left(\sqrt{x-4}+2\right)^2}+\sqrt{\left(\sqrt{x-4}-2\right)^2}=\text{|}\sqrt{x-4}+2\text{|}+\text{|}\sqrt{x-4}-2\text{|}=2\sqrt{x-4}\)

a: \(=x^2-36-x^2-14x-49+14x=-85\)

b: \(=\dfrac{5x+35+4x-28-5x-7}{\left(x-7\right)\left(x+7\right)}=\dfrac{4x}{x^2-49}\)

\(a,\left(x+6\right)\left(x-6\right)-\left(x+7\right)^2+14x=x^2-36-x^2-14x-49+14x=-85\\ b,\dfrac{5}{x-7}+\dfrac{4}{x+7}+\dfrac{5x+7}{49-x^2}=\dfrac{5\left(x+7\right)+4\left(x-7\right)-\left(5x+7\right)}{\left(x-7\right)\left(x+7\right)}=\dfrac{5x+35+4x-28-5x-7}{\left(x-7\right)\left(x+7\right)}=\dfrac{4x}{\left(x-7\right)\left(x+7\right)}\)

a)\(\left(x2+7\right).\left(x2-49\right)< 0\)

\(\left(x2+7\right).\left(x2-49\right)< 0\) chứng tỏ hai vế \(\left(x2+7\right)\) và \(\left(x2-49\right)\) khác dấu nhau .

\(\left\{{}\begin{matrix}\left(x2+7\right)>0\\\left(x2-49\right)< 0\end{matrix}\right.\)

Vì \(\left(x2+7\right)\) > \(\left(x2-49\right)\)

Nên ta có:

\(\left\{{}\begin{matrix}\left(x2+7\right)>0\\\left(x2-49\right)< 0\end{matrix}\right.\)\(\Rightarrow\)\(\left\{{}\begin{matrix}\left(x+7\right)=0\\\left(x-49\right)=0\end{matrix}\right.\)\(\Rightarrow\)\(\left\{{}\begin{matrix}x=-7\\x=49\end{matrix}\right.\)

Vậy hai số nguyên đó là -7 và 49 .

Còn phần còn lại bạn làm tương tự nhé !

!!!!!!!!!!!!!!!!!!!.............................

\(\left(2x-1\right)\left(x+7\right)=x^2-49\)

\(\Leftrightarrow\left(2x-1\right)\left(x+7\right)=\left(x-7\right)\left(x+7\right)\)

\(\Leftrightarrow\left(x+7\right)\left(x+6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-7\\x=-6\end{matrix}\right.\)

(2x-1)(x+7)=\(x^2\) -49

=> (2x-1)(x+7)=(x-7)(x+7)

=> (2x-1)(x+7)-(x-7)(x+7)=0

=>(2x-1-x+7)(x+7)=0

=> x+6=0 hoặc x+7=0

=> x=-6 hoặc x=-7