a, Xem lại đề:

b, \(16x^2-\left(4x-5\right)^2=15\)

\(\Rightarrow16x^2-\left(16x^2-40x+25\right)=15\)

\(\Rightarrow16x^2-16x^2+40x-25=15\)

\(\Rightarrow40x=40\Rightarrow x=1\)

Chúc bạn học tốt!!!

\(a.\:\left(7x+3\right)^2-4\left(x-1\right)\left(x+1\right)=49\\ 49x^2+42x+9-4x^2+4=49\\ 45x^2+42x+13=49\\ x^2+\dfrac{42}{45}x+\dfrac{13}{45}=\dfrac{49}{45}\\ x^2+2.\dfrac{7}{15}x+\left(\dfrac{7}{15}\right)^2=\dfrac{49}{45}-\dfrac{13}{45}+\left(\dfrac{7}{15}\right)^2\\ \left(x+\dfrac{7}{15}\right)^2=\dfrac{229}{225}\\ \Rightarrow\left[{}\begin{matrix}x+\dfrac{7}{15}=\dfrac{229}{225}\\x+\dfrac{7}{15}=-\dfrac{229}{225}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{124}{225}\\x=-\dfrac{334}{225}\end{matrix}\right.\)

\(c.\:\left(3x+4\right)^2-\left(3x+1\right)\left(3x-1\right)\\ =9x^2+24x+16-9x^2+1\\ 40x=-1\\ x=-\dfrac{1}{40}\)

\(d.\:\left(3x-1\right)^2-\left(3x-2\right)^2=0\\ \left(3x-1+3x-2\right)\left(3x-1-3x+2\right)=0\\ \left(6x-3\right)=0\\ x=\dfrac{1}{2}\)

\(g.\:\left(2x+1\right)^2-\left(x-1\right)^2=0\\ \left(2x+1+x-1\right)\left(2x+1-x+1\right)=0\\ 3x\left(x+2\right)=0\\ \Rightarrow\left[{}\begin{matrix}3x=0\\x+2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\)

c,\(\left(3x+4\right)^2-\left(3x-1\right)\left(3x+1\right)=49\)

\(\Rightarrow9x^2+24x+16-\left(9x^2-1\right)=49\)

\(\Rightarrow9x^2+24x+16-9x^2+1=49\)

\(\Rightarrow24x=49-1-16\)

\(\Rightarrow24x=32\Rightarrow x=\dfrac{4}{3}\)

d, \(\left(3x-1\right)^2-\left(3x-2\right)^2=0\)

\(\Rightarrow\left(3x-1-3x+2\right).\left(3x-1+3x-2\right)=0\)

\(\Rightarrow6x-3=0\Rightarrow6x=3\Rightarrow x=\dfrac{1}{2}\)

e, \(\left(2x+1\right)^2-\left(x-1\right)^2=0\)

\(\Rightarrow\left(2x+1-x+1\right)\left(2x+1+x-1\right)=0\)

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

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

Chúc bạn học tốt!!!

Tìm x biết:

b/\(\left(2x+3\right)^2-\left(5x-4\right)\left(5x+4\right)=\left(x+5\right)^2-\left(3x-1\right)\left(7x+2\right)-\left(x^2-x+1\right)\)

<=> \(4x^2 +12x+9-25x^2+16-x^2-10x-25+21x^2+6x-7x-2+x^2-x+1=0\)

<=>0x-1=0

<=>0x=1 (vô lí) (dòng này không cần ghi thêm cũng được)

=> Không có giá trị x nào thỏa mãn

c/ \((1-3x)^2-(x-2)(9x+1)=(3x-4)(3x+4)-9(x+3)^2\)

<=>\(1-6x+9x^2-9x^2-x+18x+2-9x^2+16+9x^2+54x+81=0\)

<=> 65x+100=0

<=> x=\(\dfrac{-20}{13}\)

d/\((3x+4)(3x-4)-(2x+5)^2=(x-5)^2+(2x+1)^2-(x^2-2x)+(x-1)^2\)

<=> \(9x^2-16-4x^2-20x-25-x^2+10x-25-4x^2-4x-1+x^2+2x-x^2+2x-1=0\)

<=> -10x-68=0

<=> x=\(\dfrac{-34}{5}\)

a) \(\left(4x-1\right)^2-\left(3x+2\right)\left(3x-2\right)=\left(7x-1\right)\left(x+2\right)+\left(2x+1\right)^2-\left(4x^2+7\right)\)(1)

\(\Leftrightarrow\left(16x^2-8x+1\right)-\left(9x^2-4\right)=\left(7x^2+14x-x-2\right)+\left(4x^2+4x+1\right)-\left(4x^2+7\right)\)

\(\Leftrightarrow16x^2-8x+1-9x^2+4=7x^2+13x-2+4x^2+4x+1-4x^2-7\)

\(\Leftrightarrow7x^2-8x+5=7x^2+17x-8\)

\(\Leftrightarrow7x^2-8x-7x^2-17x=-8-5\)

\(\Leftrightarrow-25x=-13\)

\(\Leftrightarrow x=\dfrac{13}{25}\)

Vậy tập nghiệm phương trình (1) là \(S=\left\{\dfrac{13}{25}\right\}\)

gắp cái gì

Tìm x

a) Ta có: \(16x^2-\left(4x-5\right)^2=15\)

\(\Leftrightarrow16x^2-\left(16x^2-40x+25\right)-15=0\)

\(\Leftrightarrow16x^2-16x^2+40x-25-15=0\)

\(\Leftrightarrow40x-40=0\)

\(\Leftrightarrow40x=40\)

hay x=1

Vậy: x=1

b) Ta có: \(\left(2x+3\right)^2-4\left(x-1\right)\left(x+1\right)=49\)

\(\Leftrightarrow4x^2+12x+9-4\left(x^2-1\right)-49=0\)

\(\Leftrightarrow4x^2+12x+9-4x^2+4-49=0\)

\(\Leftrightarrow12x-36=0\)

\(\Leftrightarrow12x=36\)

hay x=3

Vậy: x=3

d) Ta có: \(2\left(x+1\right)^2-\left(x-3\right)\left(x+3\right)-\left(x-4\right)^2=0\)

\(\Leftrightarrow2\left(x^2+2x+1\right)-\left(x^2-9\right)-\left(x^2-8x+16\right)=0\)

\(\Leftrightarrow2x^2+4x+2-x^2+9-x^2+8x-16=0\)

\(\Leftrightarrow12x-5=0\)

\(\Leftrightarrow12x=5\)

hay \(x=\frac{5}{12}\)

Vậy: \(x=\frac{5}{12}\)

e) Ta có: \(\left(x-5\right)^2-x\left(x-4\right)=9\)

\(\Leftrightarrow x^2-10x+25-x^2+4x-9=0\)

\(\Leftrightarrow-6x+16=0\)

\(\Leftrightarrow6x=16\)

hay \(x=\frac{8}{3}\)

Vậy: \(x=\frac{8}{3}\)

f) Ta có: \(\left(x-5\right)^2-\left(x-4\right)\left(1-x\right)=0\)

\(\Leftrightarrow x^2-10x+25-\left(x-x^2-4+4x\right)=0\)

\(\Leftrightarrow x^2-10x+25-x+x^2+4-4x=0\)

\(\Leftrightarrow2x^2-15x+29=0\)

\(\Leftrightarrow2\left(x^2-\frac{15}{2}x+\frac{29}{2}\right)=0\)

\(\Leftrightarrow x^2-2\cdot x\cdot\frac{15}{4}+\frac{225}{16}+\frac{7}{16}=0\)

\(\Leftrightarrow\left(x-\frac{15}{4}\right)^2+\frac{7}{16}=0\)(vô lý)

Vậy: x∈∅

\(1+\sqrt{x^2-4x+3}-x=0\)

\(ĐK:\left\{{}\begin{matrix}\sqrt{x^2-4x+3\ge0}\\x-1\ge0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x\ge3\end{matrix}\right.\)

\(PT\Leftrightarrow\sqrt{x^2-4x+3}-\left(x-1\right)=0\)

\(\Leftrightarrow\frac{x^2-4x+3-\left(x-1\right)^2}{\sqrt{x^2-4x+3}+\left(x-1\right)}=0\)

\(\Leftrightarrow2-2x=0\Rightarrow x=1\left(tm\right)\)

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

\(B\left(x\right)=3x^3+2x^2-x+5\)

b) \(A\left(x\right)-B\left(x\right)=\left(2x^3-3x^2+2x+1\right)-\left(3x^3+2x^2-x+5\right)\)

\(A\left(x\right)-B\left(x\right)=2x^3-3x^3-3x^2-2x^2+2x+x+1-5\)

\(A\left(x\right)-B\left(x\right)=-x^3-5x^2+3x-4\)

a) A(x) = 2x³- 3x² + 2x + 1

           B(x) = 3x³ + 2x² - x + 5

b) \(-\frac{2x^3-3x^2+2x+1}{3x^3+2x^2-x+5}\)= -x³ - 5x² + 3x - 4

A(x)=2x^3+2x-3x^2+1

B(x)=2x^2+3x^3-x+5

a) A(x)=2x^3-3x^2+2x+1

    B(x)=3x^3+2x^2-x+5

b)

A(x)-B(x)=2x^3-3x^2+2x+1-3x^3+2x^2-x+5=-x^3-5x^2+3x-4

Nhớ k nha